id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-001501 | Combinatorics: Counting — Permute Then Divide | 1 | Solve with verification: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 5 people from 184 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 184\\cdot(183)\\cdots(180)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1663834536}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{184}{5}$, so both methods count the same set of committees and produce 1663834536.",
"robustness_analysis": "If... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{1663834536}$.) |
math-001502 | Combinatorics: Counting — Order vs Unordered | 1 | Explain each transformation: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 2 people from 44 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by ... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{946}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{44}{2}$, so both methods count the same set of committees and produce 946.",
"robustness_analysis": "Sensitivity analysis: Perm... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{946}$.) |
math-001503 | Combinatorics: Subsets — Binomial Coefficients | 1 | Solve with verification: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 8 people from 91 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{84986896995}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{91}{8}$, so both methods count the same set of committees and produce 84986896995.",
"robustness_analysis": "Generality note: Permute-th... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001504 | Combinatorics: Counting Models — Subset Interpretation | 1 | Give a theorem-based solution: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 11 people from 25 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Expl... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4457400}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{25}{11}$, so both methods count the same set of committees and produce 4457400.",
"robustness_analysis": "Robustness note: ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001505 | Combinatorics: Counting — Permute Then Divide | 1 | Be explicit about assumptions: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 4 people from 34 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Expla... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 34\\cdot(33)\\cdots(31)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{46376}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{34}{4}$, so both methods count the same set of committees and produce 46376.",
"robustness_analysis": "Generality note: Permute-then-divide works wh... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001506 | Discrete Math: Choosing without Replacement | 1 | Give a theorem-based solution: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 12 people from 15 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing ... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 15\\cdot(14)\\cdots(4)$.",
"Step 2: Each unordered committee corresponds to... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{455}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{15}{12}$, so both methods count the same set of committees and produce 455.",
"robustness_analysis": "If the problem were perturbed: Permute-the... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001507 | Combinatorics: Counting — Order vs Unordered | 1 | Solve with verification: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 4 people from 15 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 15\\cdot(14)\\cdots(12)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1365}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{15}{4}$, so both methods count the same set of committees and produce 1365.",
"robustness_analysis": "Robustness note: Permute-then-divide work... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{1365}$.) |
math-001508 | Combinatorics: Counting Models — Subset Interpretation | 1 | Start by stating any domain restrictions: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 3 people from 94 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 94\\cdot(93)\\cdots(92)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{134044}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{94}{3}$, so both methods count the same set of committees and produce 134044.",
"robustness_analysis": "Sensitivity analysis: Permute-then-divide w... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{134044}$.) |
math-001509 | Combinatorics: Counting Models — Subset Interpretation | 1 | Show all reasoning: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 12 people from 46 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain i... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{38910617655}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{46}{12}$, so both methods count the same set of committees and produce 38910617655.",
"robustness_analysis": "Sensitivity analysis: Permute-th... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{38910617655}$.) |
math-001510 | Combinatorics: Counting — Order vs Unordered | 1 | Give a fully justified solution: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 9 people from 142 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Ex... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 142\\cdot(141)\\cdots(134)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{49941822741810}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{142}{9}$, so both methods count the same set of committees and produce 49941822741810.",
"robustness_analysis": "Generality note: Per... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{49941822741810}$.) |
math-001511 | Combinatorics: Subsets — Binomial Coefficients | 1 | Answer using clear logical steps: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 10 people from 51 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividi... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 51\\cdot(50)\\cdots(42)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{12777711870}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{51}{10}$, so both methods count the same set of committees and produce 12777711870.",
"robustness_analysis": "If the pr... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001512 | Combinatorics: Subsets — Binomial Coefficients | 1 | Answer using clear logical steps: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 10 people from 157 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and divid... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1871392332785690}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{157}{10}$, so both methods count the same set of committees and produce 1871392332785690.",
"robustness_analysis": "If the problem ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001513 | Discrete Math: Choosing without Replacement | 1 | Answer with a short justification: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 3 people from 30 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) E... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4060}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{30}{3}$, so both methods count the same set of committees and produce 4060.",
"robustness_analysis": "Sensitivity analysis: Pe... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001514 | Combinatorics: Counting — Permute Then Divide | 1 | Question: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 5 people from 150 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain ... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{591600030}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{150}{5}$, so both methods count the same set of committees and produce 591600030.",
"robustness_analysis": "Generality no... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{591600030}$.) |
math-001515 | Discrete Math: Choosing without Replacement | 1 | Challenge: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 7 people from 195 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 195\\cdot(194)\\cdots(189)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1907713332720}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{195}{7}$, so both methods count the same set of committees and produce 1907713332720.",
"robustness_analysis": "Sensitivity analysis: ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{1907713332720}$.) |
math-001516 | Combinatorics: Committees — Combinations | 1 | Solve and include a self-check: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 2 people from 22 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{231}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{22}{2}$, so both methods count the same set of committees and produce 231.",
"robustness_analysis": "Generality note: Permute-then-divide works when e... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{231}$.) |
math-001517 | Combinatorics: Counting — Order vs Unordered | 1 | Explain each transformation: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 4 people from 122 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explai... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{8783390}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{122}{4}$, so both methods count the same set of committees and produce 8783390.",
"robustness_analysis": "Robustness note: Permute-then-divide wor... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{8783390}$.) |
math-001518 | Combinatorics: Subsets — Binomial Coefficients | 1 | Give a theorem-based solution: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 2 people from 102 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5151}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{102}{2}$, so both methods count the same set of committees and produce 5151.",
"robustness_analysis": "Sensitivity analysis: Permute-then-divid... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{5151}$.) |
math-001519 | Discrete Math: Choosing without Replacement | 1 | Answer with a short justification: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 6 people from 97 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) E... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{988172368}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{97}{6}$, so both methods count the same set of committees and produce 988172368.",
"robustness_analysis": "Robustness note: Permute-then-d... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{988172368}$.) |
math-001520 | Combinatorics: Committees — Combinations | 1 | Provide both a computational and a conceptual explanation: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 3 people from 137 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections an... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 137\\cdot(136)\\cdots(135)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{419220}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{137}{3}$, so both methods count the same set of committees and produce 419220.",
"robustness_analysis": "Sensitivity analysis: Permute-then-divide ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001521 | Combinatorics: Counting — Permute Then Divide | 1 | Give reasoning, not just computation: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 3 people from 132 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 132\\cdot(131)\\cdots(130)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{374660}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{132}{3}$, so both methods count the same set of committees and produce 374660.",
"robustness_analysis": "Robustness note: Pe... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{374660}$.) |
math-001522 | Combinatorics: Counting — Permute Then Divide | 1 | Task: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 6 people from 187 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sentence... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{54768908194}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{187}{6}$, so both methods count the same set of committees and produce 54768908194.",
"robustness_analysis": "Generality note: Permute-t... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{54768908194}$.) |
math-001523 | Discrete Math: Choosing without Replacement | 1 | Proceed methodically: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 9 people from 76 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered se... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{142466675900}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{76}{9}$, so both methods count the same set of committees and produce 142466675900.",
"robustness_analysis": ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{142466675900}$.) |
math-001524 | Combinatorics: Counting — Permute Then Divide | 1 | Give an answer and a quick verification: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 12 people from 124 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing ... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 124\\cdot(123)\\cdots(113)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{15924662409664151}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{124}{12}$, so both methods count the same set of committees and produce 15924662409664151.",
"robustness_analysis": "Robustness note: Pe... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{15924662409664151}$.) |
math-001525 | Combinatorics: Subsets — Binomial Coefficients | 1 | Start by stating any domain restrictions: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 6 people from 49 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 49\\cdot(48)\\cdots(44)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{13983816}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{49}{6}$, so both methods count the same set of committees and produce 13983816.",
"robustness_analysis": "Robustness note: Permute-then-divide wo... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{13983816}$.) |
math-001526 | Combinatorics: Committees — Combinations | 1 | State any required conditions first: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 2 people from 136 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and div... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 136\\cdot(135)\\cdots(135)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{9180}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{136}{2}$, so both methods count the same set of committees and produce 9180.",
"robustness_analysis": "Sensitivity analysis: Permute-then-divid... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{9180}$.) |
math-001527 | Combinatorics: Committees — Combinations | 1 | Do not skip justification steps: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 3 people from 181 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Ex... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 181\\cdot(180)\\cdots(179)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{971970}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{181}{3}$, so both methods count the same set of committees and produce 971970.",
"robustness_analysis": "Sensitivit... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001528 | Combinatorics: Committees — Combinations | 1 | Start by stating any domain restrictions: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 11 people from 19 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve b... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{75582}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{19}{11}$, so both methods count the same set of committees and produce 75582.",
"robustness_analysis": "Robustness note: Permute-then-divide works w... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{75582}$.) |
math-001529 | Combinatorics: Committees — Combinations | 1 | Give an answer and a quick verification: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 5 people from 135 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing b... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{346700277}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{135}{5}$, so both methods count the same set of committees and produce 346700277.",
"robustness_analysis": "If t... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{346700277}$.) |
math-001530 | Combinatorics: Counting Models — Subset Interpretation | 1 | Make each step logically reversible (or explain if not): Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 4 people from 53 distinct people?
(a) Solve using the binomial coefficient definitio... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{292825}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{53}{4}$, so both methods count the same set of committees and produce 292825.",
"robustness_analysis": "Generality note: Per... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{292825}$.) |
math-001531 | Discrete Math: Choosing without Replacement | 1 | Answer using clear logical steps: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 7 people from 34 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5379616}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{34}{7}$, so both methods count the same set of committees and produce 5379616.",
"robustness_analysis": "If the problem wer... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{5379616}$.) |
math-001532 | Combinatorics: Counting — Order vs Unordered | 1 | Explain why your operations are valid: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 12 people from 174 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by ... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1090856963244211111}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{174}{12}$, so both methods count the same set of committees and produce 1090856963244211111.",
"robustness_analysis": "Robustness note... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{1090856963244211111}$.) |
math-001533 | Combinatorics: Counting — Order vs Unordered | 1 | Be explicit about assumptions: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 4 people from 179 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{41356876}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{179}{4}$, so both methods count the same set of committees and produce 41356876.",
"robustness_analysis": "Generality note: Permute-then-divide w... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{41356876}$.) |
math-001534 | Discrete Math: Choosing without Replacement | 1 | Write the solution set clearly: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 4 people from 25 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 25\\cdot(24)\\cdots(22)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{12650}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{25}{4}$, so both methods count the same set of committees and produce 12650.",
"robustness_analysis": "Generality no... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{12650}$.) |
math-001535 | Discrete Math: Choosing without Replacement | 1 | Make each step logically reversible (or explain if not): Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 12 people from 92 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selection... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 92\\cdot(91)\\cdots(81)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{362827605867363}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{92}{12}$, so both methods count the same set of committees and produce 362827605867363.",
"robustness_anal... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{362827605867363}$.) |
math-001536 | Combinatorics: Counting Models — Subset Interpretation | 1 | Solve (and briefly cross-validate): Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 9 people from 53 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 53\\cdot(52)\\cdots(45)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4431613550}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{53}{9}$, so both methods count the same set of committees and produce 4431613550.",
"robustness_analysis": "Robustness note: Permute-then-divid... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{4431613550}$.) |
math-001537 | Combinatorics: Counting — Permute Then Divide | 1 | Solve and justify each step: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 7 people from 44 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{38320568}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{44}{7}$, so both methods count the same set of committees and produce 38320568.",
"robustness_analysis": "Robustness note: Permute-then-div... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001538 | Combinatorics: Committees — Combinations | 1 | Give a fully justified solution: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 2 people from 38 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 38\\cdot(37)\\cdots(37)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{703}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{38}{2}$, so both methods count the same set of committees and produce 703.",
"robustness_analysis": "Robustness note: Permute-then-divide works when e... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001539 | Combinatorics: Counting — Permute Then Divide | 1 | Solve with verification: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 3 people from 172 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Expl... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 172\\cdot(171)\\cdots(170)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{833340}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{172}{3}$, so both methods count the same set of committees and produce 833340.",
"robustness_analysis": "Generality note: Pe... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001540 | Combinatorics: Counting — Permute Then Divide | 1 | Be explicit about assumptions: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 6 people from 43 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting o... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6096454}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{43}{6}$, so both methods count the same set of committees and produce 6096454.",
"robustness_analysis": "Robustness note: Permute-then-divide work... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001541 | Discrete Math: Choosing without Replacement | 1 | Start by stating any domain restrictions: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 8 people from 61 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 61\\cdot(60)\\cdots(54)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2944827765}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{61}{8}$, so both methods count the same set of committees and produce 2944827765.",
"robustness_analysis": "Sensitivity analysis: Permute... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001542 | Combinatorics: Counting Models — Subset Interpretation | 1 | Write the solution set clearly: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 10 people from 22 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{646646}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{22}{10}$, so both methods count the same set of committees and produce 646646.",
"robustness_analysis": "Generality... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{646646}$.) |
math-001543 | Combinatorics: Counting — Permute Then Divide | 1 | Work carefully and justify each inference: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 2 people from 109 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 109\\cdot(108)\\cdots(108)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5886}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{109}{2}$, so both methods count the same set of committees and produce 5886.",
"robustness_analysis": "Sensitivity an... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{5886}$.) |
math-001544 | Combinatorics: Counting — Order vs Unordered | 1 | Carefully track domains: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 12 people from 123 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 123\\cdot(122)\\cdots(112)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{14383566047438588}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{123}{12}$, so both methods count the same set of committees and produce 14383566047438588.",
"robustness_analysis": "If the proble... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{14383566047438588}$.) |
math-001545 | Combinatorics: Subsets — Binomial Coefficients | 1 | Indicate where a theorem is used: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 5 people from 22 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividin... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{26334}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{22}{5}$, so both methods count the same set of committees and produce 26334.",
"robustness_analysis": "If the problem were perturbed: Permute-... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001546 | Discrete Math: Choosing without Replacement | 1 | Complete the analysis: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 5 people from 186 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explai... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 186\\cdot(185)\\cdots(182)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1757291172}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{186}{5}$, so both methods count the same set of committees and produce 1757291172.",
"robustness_analysis": "Ge... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001547 | Combinatorics: Counting — Permute Then Divide | 1 | Try to avoid pattern-matching; explain why: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 12 people from 77 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections ... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 77\\cdot(76)\\cdots(66)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{36749279048405}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{77}{12}$, so both methods count the same set of committees and produce 36749279048405.",
"robustness_analysis": "If the problem were ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001548 | Combinatorics: Counting Models — Subset Interpretation | 1 | Give reasoning, not just computation: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 2 people from 106 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by co... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 106\\cdot(105)\\cdots(105)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5565}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{106}{2}$, so both methods count the same set of committees and produce 5565.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{5565}$.) |
math-001549 | Discrete Math: Choosing without Replacement | 1 | Proceed methodically: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 6 people from 108 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1913554188}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{108}{6}$, so both methods count the same set of committees and produce 1913554188.",
"robustness_analysis": "Robustness note: Permute-then-divi... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001550 | Discrete Math: Choosing without Replacement | 1 | Carefully track domains: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 3 people from 35 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in ... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6545}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{35}{3}$, so both methods count the same set of committees and produce 6545.",
"robustness_analysis": "If the problem were pert... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{6545}$.) |
math-001551 | Combinatorics: Committees — Combinations | 1 | Task: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 9 people from 15 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sentence why di... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 15\\cdot(14)\\cdots(7)$.",
"Step 2: Each unordered committee corresponds to... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5005}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{15}{9}$, so both methods count the same set of committees and produce 5005.",
"robustness_analysis": "Sensitivity analysis: Pe... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001552 | Discrete Math: Choosing without Replacement | 1 | Prompt: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 9 people from 20 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sentence why ... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 20\\cdot(19)\\cdots(12)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{167960}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{20}{9}$, so both methods count the same set of committees and produce 167960.",
"robustness_analysis": "Generality ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001553 | Combinatorics: Counting — Order vs Unordered | 1 | Carefully track domains: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 6 people from 158 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{19628752143}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{158}{6}$, so both methods count the same set of committees and produce 19628752143.",
"robustness_analysis": "Sensitivity analysis: Perm... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{19628752143}$.) |
math-001554 | Combinatorics: Counting — Order vs Unordered | 1 | Explain what is being counted/optimized: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 8 people from 146 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing b... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4212589656990}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{146}{8}$, so both methods count the same set of committees and produce 4212589656990.",
"robustness_analysis": "If th... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{4212589656990}$.) |
math-001555 | Combinatorics: Counting Models — Subset Interpretation | 1 | State any required conditions first: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 9 people from 54 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and divi... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 54\\cdot(53)\\cdots(46)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5317936260}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{54}{9}$, so both methods count the same set of committees and produce 5317936260.",
"robustness_analysis": "Robustness n... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001556 | Combinatorics: Committees — Combinations | 1 | Solve and then verify: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 3 people from 96 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in on... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{142880}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{96}{3}$, so both methods count the same set of committees and produce 142880.",
"robustness_analysis": "If the problem were perturbed: Permut... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{142880}$.) |
math-001557 | Combinatorics: Subsets — Binomial Coefficients | 1 | Proceed methodically: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 3 people from 145 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in on... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 145\\cdot(144)\\cdots(143)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{497640}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{145}{3}$, so both methods count the same set of committees and produce 497640.",
"robustness_analysis": "Robustness... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{497640}$.) |
math-001558 | Discrete Math: Choosing without Replacement | 1 | Answer using clear logical steps: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 4 people from 51 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Ex... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{249900}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{51}{4}$, so both methods count the same set of committees and produce 249900.",
"robustness_analysis": "Sensitivity analysis... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{249900}$.) |
math-001559 | Combinatorics: Counting — Permute Then Divide | 1 | Give an answer and a quick verification: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 10 people from 163 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing ... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 163\\cdot(162)\\cdots(154)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2753064116158356}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{163}{10}$, so both methods count the same set of committees and produce 2753064116158356.",
"robustness_analysis": "Robustness note... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001560 | Combinatorics: Counting Models — Subset Interpretation | 1 | Explain what is being counted/optimized: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 7 people from 58 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and ... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{300674088}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{58}{7}$, so both methods count the same set of committees and produce 300674088.",
"robustness_analysis": "Sensitivity analysis: Permute-then-di... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{300674088}$.) |
math-001561 | Combinatorics: Counting Models — Subset Interpretation | 1 | Give a fully justified solution: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 5 people from 85 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{32801517}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{85}{5}$, so both methods count the same set of committees and produce 32801517.",
"robustness_analysis": "Sensitivity analysis: Permute-the... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001562 | Combinatorics: Committees — Combinations | 1 | Solve and justify each step: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 2 people from 171 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) ... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{14535}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{171}{2}$, so both methods count the same set of committees and produce 14535.",
"robustness_analysis": "If the problem were perturbed: Permute-then-... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001563 | Combinatorics: Counting Models — Subset Interpretation | 1 | Indicate where a theorem is used: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 2 people from 93 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by countin... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 93\\cdot(92)\\cdots(92)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4278}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{93}{2}$, so both methods count the same set of committees and produce 4278.",
"robustness_analysis": "If the problem were pert... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{4278}$.) |
math-001564 | Combinatorics: Counting Models — Subset Interpretation | 1 | Task: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 9 people from 105 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sentence... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 105\\cdot(104)\\cdots(97)$.",
"Step 2: Each unordered committee corresponds... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3005047770725}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{105}{9}$, so both methods count the same set of committees and produce 3005047770725.",
"robustness_analysis": "Robustness note: Permute-the... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{3005047770725}$.) |
math-001565 | Combinatorics: Committees — Combinations | 1 | Write the solution set clearly: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 4 people from 168 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 168\\cdot(167)\\cdots(165)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{32018910}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{168}{4}$, so both methods count the same set of committees and produce 32018910.",
"robustness_analysis": "Robustness note... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{32018910}$.) |
math-001566 | Discrete Math: Choosing without Replacement | 1 | Solve and sanity-check: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 4 people from 124 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{9381251}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{124}{4}$, so both methods count the same set of committees and produce 9381251.",
"robustness_analysis": "Sensitiv... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{9381251}$.) |
math-001567 | Combinatorics: Subsets — Binomial Coefficients | 1 | Checkpoint: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 10 people from 198 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 198\\cdot(197)\\cdots(189)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{20256672480820776}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{198}{10}$, so both methods count the same set of committees and produce 20256672480820776.",
"robustness_analysis": "If the proble... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001568 | Combinatorics: Counting — Permute Then Divide | 1 | Proceed methodically: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 10 people from 152 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 152\\cdot(151)\\cdots(143)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1340705736329960}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{152}{10}$, so both methods count the same set of committees and produce 1340705736329960.",
"robustness_analysis":... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001569 | Combinatorics: Subsets — Binomial Coefficients | 1 | Exercise: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 12 people from 86 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sent... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 86\\cdot(85)\\cdots(75)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{152902266958350}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{86}{12}$, so both methods count the same set of committees and produce 152902266958350.",
"robustness_analysis": "G... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001570 | Combinatorics: Counting — Permute Then Divide | 1 | Task: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 12 people from 132 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in ... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 132\\cdot(131)\\cdots(121)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{34898565177533200}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{132}{12}$, so both methods count the same set of committees and produce 34898565177533200.",
"robustness_analysis": "If the proble... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{34898565177533200}$.) |
math-001571 | Combinatorics: Counting — Order vs Unordered | 1 | Exercise: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 9 people from 35 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain i... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 35\\cdot(34)\\cdots(27)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{70607460}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{35}{9}$, so both methods count the same set of committees and produce 70607460.",
"robustness_analysis": "If the ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001572 | Combinatorics: Subsets — Binomial Coefficients | 1 | Try to avoid pattern-matching; explain why: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 9 people from 92 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{868754947060}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{92}{9}$, so both methods count the same set of committees and produce 868754947060.",
"robustness_analysis": "Sensitiv... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{868754947060}$.) |
math-001573 | Combinatorics: Committees — Combinations | 1 | Complete the analysis: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 11 people from 88 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered ... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{32006008361808}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{88}{11}$, so both methods count the same set of committees and produce 32006008361808.",
"robustness_analysis": "If the problem were pertur... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{32006008361808}$.) |
math-001574 | Combinatorics: Subsets — Binomial Coefficients | 1 | Do not skip justification steps: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 9 people from 172 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{293466669659020}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{172}{9}$, so both methods count the same set of committees and produce 293466669659020.",
"robustness_analysis": "Robustness note: P... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{293466669659020}$.) |
math-001575 | Combinatorics: Counting — Order vs Unordered | 1 | Give an answer and a quick verification: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 10 people from 200 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections an... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{22451004309013280}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{200}{10}$, so both methods count the same set of committees and produce 22451004309013280.",
"robustness_analysis": "Generality no... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{22451004309013280}$.) |
math-001576 | Combinatorics: Committees — Combinations | 1 | Do not skip justification steps: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 2 people from 128 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Ex... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 128\\cdot(127)\\cdots(127)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{8128}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{128}{2}$, so both methods count the same set of committees and produce 8128.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001577 | Combinatorics: Committees — Combinations | 1 | Explain each transformation: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 10 people from 183 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Expla... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{9037676445227430}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{183}{10}$, so both methods count the same set of committees and produce 9037676445227430.",
"robustness_analysis": "If the problem were p... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{9037676445227430}$.) |
math-001578 | Combinatorics: Committees — Combinations | 1 | Warm-up: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 9 people from 70 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sentence why... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 70\\cdot(69)\\cdots(62)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{65033528560}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{70}{9}$, so both methods count the same set of committees and produce 65033528560.",
"robustness_analysis": "G... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{65033528560}$.) |
math-001579 | Combinatorics: Counting — Order vs Unordered | 1 | Proceed methodically: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 4 people from 180 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered s... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{42296805}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{180}{4}$, so both methods count the same set of committees and produce 42296805.",
"robustness_analysis": "Generality note... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{42296805}$.) |
math-001580 | Combinatorics: Counting — Order vs Unordered | 1 | Make each step logically reversible (or explain if not): How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 8 people from 70 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{9440350920}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{70}{8}$, so both methods count the same set of committees and produce 9440350920.",
"robustness_analysis": "Sensitivity analysis: Permute-then-... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{9440350920}$.) |
math-001581 | Combinatorics: Counting — Permute Then Divide | 1 | Derive the result step-by-step: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 12 people from 51 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 51\\cdot(50)\\cdots(40)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{158753389900}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{51}{12}$, so both methods count the same set of committees and produce 158753389900.",
"robustness_analysis":... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{158753389900}$.) |
math-001582 | Discrete Math: Choosing without Replacement | 1 | Work this out carefully: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 4 people from 109 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 109\\cdot(108)\\cdots(106)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5563251}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{109}{4}$, so both methods count the same set of committees and produce 5563251.",
"robustness_analysis": "Sensitiv... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{5563251}$.) |
math-001583 | Combinatorics: Subsets — Binomial Coefficients | 1 | Write the solution set clearly: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 4 people from 43 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Expl... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{123410}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{43}{4}$, so both methods count the same set of committees and produce 123410.",
"robustness_analysis": "Sensitivity analysis... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{123410}$.) |
math-001584 | Combinatorics: Counting Models — Subset Interpretation | 1 | Give a theorem-based solution: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 3 people from 23 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting o... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 23\\cdot(22)\\cdots(21)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1771}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{23}{3}$, so both methods count the same set of committees and produce 1771.",
"robustness_analysis": "Robustness note... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{1771}$.) |
math-001585 | Combinatorics: Counting Models — Subset Interpretation | 1 | Task: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 9 people from 18 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in on... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 18\\cdot(17)\\cdots(10)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{48620}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{18}{9}$, so both methods count the same set of committees and produce 48620.",
"robustness_analysis": "If the problem were pe... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{48620}$.) |
math-001586 | Combinatorics: Subsets — Binomial Coefficients | 1 | State any required conditions first: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 7 people from 81 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by coun... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 81\\cdot(80)\\cdots(75)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3477216600}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{81}{7}$, so both methods count the same set of committees and produce 3477216600.",
"robustness_analysis": "Robustness n... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001587 | Discrete Math: Choosing without Replacement | 1 | Problem: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 10 people from 170 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sentence w... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4241922417794061}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{170}{10}$, so both methods count the same set of committees and produce 4241922417794061.",
"robustness_analysis": "If the problem were p... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{4241922417794061}$.) |
math-001588 | Combinatorics: Counting Models — Subset Interpretation | 1 | Prompt: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 3 people from 91 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and d... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{121485}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{91}{3}$, so both methods count the same set of committees and produce 121485.",
"robustness_analysis": "If the prob... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{121485}$.) |
math-001589 | Combinatorics: Subsets — Binomial Coefficients | 1 | Challenge: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 11 people from 91 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sen... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{47325339895743}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{91}{11}$, so both methods count the same set of committees and produce 47325339895743.",
"robustness_analysis": "Sensitivity analysis... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001590 | Combinatorics: Counting — Order vs Unordered | 1 | Question: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 6 people from 93 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain i... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{762245484}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{93}{6}$, so both methods count the same set of committees and produce 762245484.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{762245484}$.) |
math-001591 | Combinatorics: Counting — Permute Then Divide | 1 | Solve and include a self-check: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 3 people from 92 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing ... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 92\\cdot(91)\\cdots(90)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{125580}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{92}{3}$, so both methods count the same set of committees and produce 125580.",
"robustness_analysis": "Robustness ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{125580}$.) |
math-001592 | Combinatorics: Counting — Permute Then Divide | 1 | Determine the requested value: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 5 people from 147 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{534017484}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{147}{5}$, so both methods count the same set of committees and produce 534017484.",
"robustness_analysis": "Robustness no... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001593 | Combinatorics: Committees — Combinations | 1 | Start by stating any domain restrictions: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 4 people from 79 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1502501}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{79}{4}$, so both methods count the same set of committees and produce 1502501.",
"robustness_analysis": "If the pr... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001594 | Combinatorics: Counting Models — Subset Interpretation | 1 | Solve and then verify: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 6 people from 183 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explai... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 183\\cdot(182)\\cdots(178)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{48019108683}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{183}{6}$, so both methods count the same set of committees and produce 48019108683.",
"robustness_analysis": "Sensitivi... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{48019108683}$.) |
math-001595 | Combinatorics: Committees — Combinations | 1 | Exercise: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 8 people from 125 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections an... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1176174344125}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{125}{8}$, so both methods count the same set of committees and produce 1176174344125.",
"robustness_analysis": "Robustness note: Permute-the... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{1176174344125}$.) |
math-001596 | Combinatorics: Counting — Permute Then Divide | 1 | Try to avoid pattern-matching; explain why: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 6 people from 160 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividin... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 160\\cdot(159)\\cdots(155)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{21193254160}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{160}{6}$, so both methods count the same set of committees and produce 21193254160.",
"robustness_analysis": "... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{21193254160}$.) |
math-001597 | Combinatorics: Counting — Order vs Unordered | 1 | Answer using clear logical steps: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 5 people from 75 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Ex... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 75\\cdot(74)\\cdots(71)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{17259390}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{75}{5}$, so both methods count the same set of committees and produce 17259390.",
"robustness_analysis": "Sensitivity analysis: Permute-then-divi... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{17259390}$.) |
math-001598 | Combinatorics: Committees — Combinations | 1 | Problem: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 7 people from 135 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sentence wh... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 135\\cdot(134)\\cdots(129)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{138432467745}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{135}{7}$, so both methods count the same set of committees and produce 138432467745.",
"robustness_analysis":... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{138432467745}$.) |
math-001599 | Combinatorics: Counting — Permute Then Divide | 1 | Give a fully justified solution: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 8 people from 185 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Ex... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{29188758206595}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{185}{8}$, so both methods count the same set of committees and produce 29188758206595.",
"robustness_analys... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001600 | Combinatorics: Counting — Permute Then Divide | 1 | Solve with verification: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 3 people from 69 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{52394}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{69}{3}$, so both methods count the same set of committees and produce 52394.",
"robustness_analysis": "Robustness note: Permute-then-divide works wh... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{52394}$.) |
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