id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-001401 | Combinatorics: Counting — Permute Then Divide | 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 3 people from 17 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by cou... | [
{
"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: $\\boxed{680}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{17}{3}$, so both methods count the same set of committees and produce 680.",
"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... | 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 $\boxed{680}$.) |
math-001402 | Combinatorics: Counting — Order vs Unordered | 1 | Show all reasoning: 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 116 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered sel... | [
{
"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: $\\boxed{46627515440}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{116}{7}$, so both methods count the same set of committees and produce 46627515440.",
"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... | 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 $\boxed{46627515440}$.) |
math-001403 | Combinatorics: Committees — Combinations | 1 | Explain each transformation: 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 176 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: $\\boxed{6057158960772920}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{176}{10}$, so both methods count the same set of committees and produce 6057158960772920.",
"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 $\boxed{6057158960772920}$.) |
math-001404 | Combinatorics: Counting Models — Subset Interpretation | 1 | State any required conditions first: 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 71 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: $\\boxed{57155}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{71}{3}$, so both methods count the same set of committees and produce 57155.",
"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... | 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 $\boxed{57155}$.) |
math-001405 | Combinatorics: Committees — Combinations | 1 | Explain each transformation: 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 5 people from 109 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting or... | [
{
"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: $\\boxed{116828271}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{109}{5}$, so both methods count the same set of committees and produce 116828271.",
"robustness_analysis": "Sens... | [
{
"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 $\boxed{116828271}$.) |
math-001406 | Discrete Math: Choosing without Replacement | 1 | Keep the final answer in boxed form: 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 11 people from 136 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and 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: $nP k = 136\\cdot(135)\\cdots(126)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4868956823342976}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{136}{11}$, so both methods count the same set of committees and produce 4868956823342976.",
"robustness_analysis": "Generality note: 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... | 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 $\boxed{4868956823342976}$.) |
math-001407 | Combinatorics: Counting — Permute Then Divide | 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 10 people from 145 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": "Both approaches agree after simplification. Final answer: $\\boxed{824313388697656}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{145}{10}$, so both methods count the same set of committees and produce 824313388697656.",
"robustness_analysis": "Robustness note: 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 $k!$'. |
math-001408 | Combinatorics: Committees — Combinations | 1 | Start by stating any domain restrictions: 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 76 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": "Both approaches agree after simplification. Final answer: $\\boxed{31022118677225}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{76}{12}$, so both methods count the same set of committees and produce 31022118677225.",
"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... | 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 $\boxed{31022118677225}$.) |
math-001409 | Combinatorics: Subsets — Binomial Coefficients | 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 12 people from 102 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: $nP k = 102\\cdot(101)\\cdots(91)$.",
"Step 2: Each unordered committee corresponds... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1350990969850340}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{102}{12}$, so both methods count the same set of committees and produce 1350990969850340.",
"robustness_a... | [
{
"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 $\boxed{1350990969850340}$.) |
math-001410 | Combinatorics: Counting Models — Subset Interpretation | 1 | Make each step logically reversible (or explain if not): 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 148 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and ... | [
{
"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: $nP k = 148\\cdot(147)\\cdots(142)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{267212177232}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{148}{7}$, so both methods count the same set of committees and produce 267212177232.",
"robustness_analysis": "Robustness note: 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... | 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 $\boxed{267212177232}$.) |
math-001411 | Combinatorics: Counting — Permute Then Divide | 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 6 people from 115 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selection... | [
{
"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: $\\boxed{2813729380}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{115}{6}$, so both methods count the same set of committees and produce 2813729380.",
"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... | 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 $\boxed{2813729380}$.) |
math-001412 | Discrete Math: Choosing without Replacement | 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 2 people from 115 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: $\\boxed{6555}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{115}{2}$, so both methods count the same set of committees and produce 6555.",
"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... | 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-001413 | Combinatorics: Counting — Permute Then Divide | 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 12 people from 109 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: $nP k = 109\\cdot(108)\\cdots(98)$.",
"Step 2: Each unordered committee corresponds... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3133614810784185}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{109}{12}$, so both methods count the same set of committees and produce 3133614810784185.",
"robustness_analysis": "Sensitivity ana... | [
{
"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 $\boxed{3133614810784185}$.) |
math-001414 | Combinatorics: Subsets — Binomial Coefficients | 1 | Prompt: 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 20 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: $nP k = 20\\cdot(19)\\cdots(11)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{184756}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{20}{10}$, so both methods count the same set of committees and produce 184756.",
"robustness_analysis": "Robustness note: Permute-then-divide works... | [
{
"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-001415 | Combinatorics: Counting — Permute Then Divide | 1 | Task: 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 5 people from 127 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and di... | [
{
"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: $\\boxed{254231775}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{127}{5}$, so both methods count the same set of committees and produce 254231775.",
"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 $\boxed{254231775}$.) |
math-001416 | Discrete Math: Choosing without Replacement | 1 | Checkpoint: 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 39 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: $nP k = 39\\cdot(38)\\cdots(37)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{9139}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{39}{3}$, so both methods count the same set of committees and produce 9139.",
"robustness_analysis": "Generality note: 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... | 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 $\boxed{9139}$.) |
math-001417 | Combinatorics: Counting Models — Subset Interpretation | 1 | Work this out carefully: 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 170 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: $\\boxed{718400660120}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{170}{7}$, so both methods count the same set of committees and produce 718400660120.",
"robustness_analysis": "Robustn... | [
{
"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-001418 | Combinatorics: Committees — Combinations | 1 | Where appropriate, name the theorem you use: 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 67 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{2211}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{67}{2}$, so both methods count the same set of committees and produce 2211.",
"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... | 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 $\boxed{2211}$.) |
math-001419 | Combinatorics: Counting Models — Subset Interpretation | 1 | Solve and sanity-check: 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 76 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in 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: $nP k = 76\\cdot(75)\\cdots(69)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{18855883575}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{76}{8}$, so both methods count the same set of committees and produce 18855883575.",
"robustness_analysis": "Sensitivity analysis: Permu... | [
{
"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 $\boxed{18855883575}$.) |
math-001420 | Combinatorics: Committees — Combinations | 1 | Answer with a short justification: 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 163 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: $\\boxed{532318800456}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{163}{7}$, so both methods count the same set of committees and produce 532318800456.",
"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... | 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 $\boxed{532318800456}$.) |
math-001421 | Discrete Math: Choosing without Replacement | 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 70 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in 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: $\\boxed{10638894058520}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{70}{12}$, so both methods count the same set of committees and produce 10638894058520.",
"robustness_analysis": "Robustness 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... | 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 $\boxed{10638894058520}$.) |
math-001422 | Combinatorics: Counting Models — Subset Interpretation | 1 | Challenge: 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 5 people from 180 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections a... | [
{
"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: $\\boxed{1488847536}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{180}{5}$, so both methods count the same set of committees and produce 1488847536.",
"robustness_analysis": "If the problem were perturbed: 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... | 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 $\boxed{1488847536}$.) |
math-001423 | Combinatorics: Committees — Combinations | 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 12 people from 104 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: $nP k = 104\\cdot(103)\\cdots(93)$.",
"Step 2: Each unordered committee corresponds... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1728597141547640}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{104}{12}$, so both methods count the same set of committees and produce 1728597141547640.",
"robustness_a... | [
{
"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 $\boxed{1728597141547640}$.) |
math-001424 | Combinatorics: Subsets — Binomial Coefficients | 1 | Answer with a short justification: 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 32 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counti... | [
{
"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: $\\boxed{4960}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{32}{3}$, so both methods count the same set of committees and produce 4960.",
"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... | 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 $\boxed{4960}$.) |
math-001425 | Combinatorics: Counting — Permute Then Divide | 1 | Problem: 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 17 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{6188}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{17}{12}$, so both methods count the same set of committees and produce 6188.",
"robustness_analysis": "If the problem were perturbed: 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... | 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-001426 | Combinatorics: Counting Models — Subset Interpretation | 1 | Solve (and briefly cross-validate): 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 119 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: $nP k = 119\\cdot(118)\\cdots(117)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{273819}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{119}{3}$, so both methods count the same set of committees and produce 273819.",
"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... | 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-001427 | 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 192 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": "Both approaches agree after simplification. Final answer: $\\boxed{64300886496}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{192}{6}$, so both methods count the same set of committees and produce 64300886496.",
"robustness_analysis": "Robustness note: 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... | 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-001428 | Combinatorics: Counting Models — Subset Interpretation | 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 5 people from 140 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: $nP k = 140\\cdot(139)\\cdots(136)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{416965528}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{140}{5}$, so both methods count the same set of committees and produce 416965528.",
"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 $k!$'. |
math-001429 | Discrete Math: Choosing without Replacement | 1 | Carefully track domains: 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 40 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: $\\boxed{780}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{40}{2}$, so both methods count the same set of committees and produce 780.",
"robustness_analysis": "Sensitivity analysis: 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... | 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-001430 | Discrete Math: Choosing without Replacement | 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 2 people from 158 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{12403}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{158}{2}$, so both methods count the same set of committees and produce 12403.",
"robustness_analysis": "Robustness note: 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... | 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-001431 | Combinatorics: Subsets — Binomial Coefficients | 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 12 people from 68 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: $nP k = 68\\cdot(67)\\cdots(57)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{7282025622664}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{68}{12}$, so both methods count the same set of committees and produce 7282025622664.",
"robustness_analysis": "Robustness note: Permu... | [
{
"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-001432 | Combinatorics: Counting — Order vs Unordered | 1 | Answer with a short justification: 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 13 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: $nP k = 13\\cdot(12)\\cdots(12)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{78}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{13}{2}$, so both methods count the same set of committees and produce 78.",
"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... | 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-001433 | Combinatorics: Counting Models — Subset Interpretation | 1 | Compute the requested quantity: 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 29 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: $\\boxed{10015005}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{29}{9}$, so both methods count the same set of committees and produce 10015005.",
"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 $k!$'. (Here the result is $\boxed{10015005}$.) |
math-001434 | Combinatorics: Counting — Permute Then Divide | 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 3 people from 13 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: $\\boxed{286}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{13}{3}$, so both methods count the same set of committees and produce 286.",
"robustness_analysis": "Robustness 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... | 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-001435 | Discrete Math: Choosing without Replacement | 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 2 people from 69 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": "Both approaches agree after simplification. Final answer: $\\boxed{2346}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{69}{2}$, so both methods count the same set of committees and produce 2346.",
"robustness_analysis": "Sensitivity analysis: Permute-then-divide works... | [
{
"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-001436 | 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 6 people from 65 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{82598880}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{65}{6}$, so both methods count the same set of committees and produce 82598880.",
"robustness_analysis": "General... | [
{
"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 $\boxed{82598880}$.) |
math-001437 | Combinatorics: Subsets — Binomial Coefficients | 1 | Indicate where a theorem is used: 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 12 people from 127 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) ... | [
{
"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: $nP k = 127\\cdot(126)\\cdots(116)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{21501728724901425}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{127}{12}$, so both methods count the same set of committees and produce 21501728724901425.",
"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 $k!$'. |
math-001438 | Combinatorics: Subsets — Binomial Coefficients | 1 | Solve and sanity-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 11 people from 197 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordere... | [
{
"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: $nP k = 197\\cdot(196)\\cdots(187)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{326971339639915152}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{197}{11}$, so both methods count the same set of committees and produce 326971339639915152.",
"robustne... | [
{
"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 $\boxed{326971339639915152}$.) |
math-001439 | Combinatorics: Subsets — Binomial Coefficients | 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 8 people from 42 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: $nP k = 42\\cdot(41)\\cdots(35)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{118030185}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{42}{8}$, so both methods count the same set of committees and produce 118030185.",
"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 $k!$'. |
math-001440 | 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 3 people from 79 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: $nP k = 79\\cdot(78)\\cdots(77)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{79079}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{79}{3}$, so both methods count the same set of committees and produce 79079.",
"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... | 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-001441 | Combinatorics: Counting — Permute Then Divide | 1 | Keep the final answer in boxed form: 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 103 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: $\\boxed{1429840335}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{103}{6}$, so both methods count the same set of committees and produce 1429840335.",
"robustness_analysis": "Generality 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 $\boxed{1429840335}$.) |
math-001442 | Combinatorics: Counting Models — Subset Interpretation | 1 | Show all reasoning: 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 41 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: $\\boxed{95548245}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{41}{8}$, so both methods count the same set of committees and produce 95548245.",
"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... | 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 $\boxed{95548245}$.) |
math-001443 | Combinatorics: Subsets — Binomial Coefficients | 1 | Track quantifiers carefully: 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 36 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ord... | [
{
"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: $nP k = 36\\cdot(35)\\cdots(35)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{630}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{36}{2}$, so both methods count the same set of committees and produce 630.",
"robustness_analysis": "Robustness 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 $k!$'. |
math-001444 | Discrete Math: Choosing without Replacement | 1 | Track units/moduli 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 5 people from 98 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: $nP k = 98\\cdot(97)\\cdots(94)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{67910864}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{98}{5}$, so both methods count the same set of committees and produce 67910864.",
"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... | 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-001445 | Combinatorics: Committees — Combinations | 1 | Solve and sanity-check: 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 57 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: $nP k = 57\\cdot(56)\\cdots(56)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1596}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{57}{2}$, so both methods count the same set of committees and produce 1596.",
"robustness_analysis": "Sensitivity analysis: Permute-then-divide works... | [
{
"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-001446 | Discrete Math: Choosing without Replacement | 1 | Keep the final answer in boxed form: 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 123 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: $\\boxed{302621}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{123}{3}$, so both methods count the same set of committees and produce 302621.",
"robustness_analysis": "Generality note: Permute-then-divide works... | [
{
"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 $\boxed{302621}$.) |
math-001447 | Combinatorics: Counting — Order vs Unordered | 1 | Try to avoid pattern-matching; explain why: 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 78 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve ... | [
{
"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: $nP k = 78\\cdot(77)\\cdots(71)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{23446881315}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{78}{8}$, so both methods count the same set of committees and produce 23446881315.",
"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... | 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 $\boxed{23446881315}$.) |
math-001448 | Combinatorics: Counting Models — Subset Interpretation | 1 | Use two approaches if possible: 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 86 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: $\\boxed{5373200880}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{86}{7}$, so both methods count the same set of committees and produce 5373200880.",
"robustness_analysis": "If the problem were perturbed: 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 $\boxed{5373200880}$.) |
math-001449 | 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 7 people from 91 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: $\\boxed{8093990190}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{91}{7}$, so both methods count the same set of committees and produce 8093990190.",
"robustness_analysis": "Robustness note: 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 $\boxed{8093990190}$.) |
math-001450 | Combinatorics: Subsets — Binomial Coefficients | 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 8 people from 175 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: $nP k = 175\\cdot(174)\\cdots(168)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{18547370252775}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{175}{8}$, so both methods count the same set of committees and produce 18547370252775.",
"robustness_analysis": "Robustness 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 $\boxed{18547370252775}$.) |
math-001451 | 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 12 people from 88 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counti... | [
{
"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: $nP k = 88\\cdot(87)\\cdots(77)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{205371886988268}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{88}{12}$, so both methods count the same set of committees and produce 205371886988268.",
"robustness_analysis": "S... | [
{
"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-001452 | Combinatorics: Counting — Order vs Unordered | 1 | Try to avoid pattern-matching; explain why: 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 65 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve ... | [
{
"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: $nP k = 65\\cdot(64)\\cdots(62)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{677040}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{65}{4}$, so both methods count the same set of committees and produce 677040.",
"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 $k!$'. |
math-001453 | Combinatorics: Counting — Permute Then Divide | 1 | Prompt: 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 85 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one senten... | [
{
"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: $\\boxed{3129162672636}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{85}{10}$, so both methods count the same set of committees and produce 3129162672636.",
"robustness_analysis": "Sensitivity analysis: 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... | 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 $\boxed{3129162672636}$.) |
math-001454 | Combinatorics: Subsets — Binomial Coefficients | 1 | Exercise: 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 165 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": "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: $nP k = 165\\cdot(164)\\cdots(159)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{580688008560}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{165}{7}$, so both methods count the same set of committees and produce 580688008560.",
"robustness_analysis": "General... | [
{
"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-001455 | Combinatorics: Committees — Combinations | 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 5 people from 137 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: $nP k = 137\\cdot(136)\\cdots(133)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{373566942}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{137}{5}$, so both methods count the same set of committees and produce 373566942.",
"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 $k!$'. |
math-001456 | Combinatorics: Counting — Order vs Unordered | 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 8 people from 122 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: $nP k = 122\\cdot(121)\\cdots(115)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{962889794295}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{122}{8}$, so both methods count the same set of committees and produce 962889794295.",
"robustness_analysis": "Robustness note: 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 $k!$'. (Here the result is $\boxed{962889794295}$.) |
math-001457 | Combinatorics: Counting — Permute Then Divide | 1 | Find the exact value: 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 159 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: $nP k = 159\\cdot(158)\\cdots(155)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{794747031}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{159}{5}$, so both methods count the same set of committees and produce 794747031.",
"robustness_analysis": "Gene... | [
{
"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 $\boxed{794747031}$.) |
math-001458 | Discrete Math: Choosing without Replacement | 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 10 people from 34 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: $\\boxed{131128140}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{34}{10}$, so both methods count the same set of committees and produce 131128140.",
"robustness_analysis": "If the problem were perturbed:... | [
{
"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 $\boxed{131128140}$.) |
math-001459 | Combinatorics: Committees — Combinations | 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 5 people from 126 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: $nP k = 126\\cdot(125)\\cdots(122)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{244222650}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{126}{5}$, so both methods count the same set of committees and produce 244222650.",
"robustness_analysis": "Sensitivity analysis: 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... | 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 $\boxed{244222650}$.) |
math-001460 | Combinatorics: Counting — Permute Then Divide | 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 11 people from 18 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Exp... | [
{
"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: $\\boxed{31824}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{18}{11}$, so both methods count the same set of committees and produce 31824.",
"robustness_analysis": "Sensitivity analysis: 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... | 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-001461 | Combinatorics: Committees — Combinations | 1 | Problem: 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 173 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sente... | [
{
"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: $nP k = 173\\cdot(172)\\cdots(169)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1218218079}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{173}{5}$, so both methods count the same set of committees and produce 1218218079.",
"robustness_analysis": "Se... | [
{
"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-001462 | Combinatorics: Counting Models — Subset Interpretation | 1 | Solve and then verify: 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 22 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered ... | [
{
"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: $nP k = 22\\cdot(21)\\cdots(12)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{705432}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{22}{11}$, so both methods count the same set of committees and produce 705432.",
"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... | 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-001463 | Combinatorics: Counting — Order vs Unordered | 1 | Question: 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 12 people from 125 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{17615777001840875}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{125}{12}$, so both methods count the same set of committees and produce 17615777001840875.",
"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-001464 | Combinatorics: Subsets — Binomial Coefficients | 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 3 people from 116 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: $nP k = 116\\cdot(115)\\cdots(114)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{253460}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{116}{3}$, so both methods count the same set of committees and produce 253460.",
"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... | 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-001465 | Combinatorics: Counting — Order vs Unordered | 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 10 people from 99 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: $\\boxed{15579278510796}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{99}{10}$, so both methods count the same set of committees and produce 15579278510796.",
"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 $k!$'. |
math-001466 | 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 10 people from 98 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) E... | [
{
"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: $nP k = 98\\cdot(97)\\cdots(89)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{14005614014756}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{98}{10}$, so both methods count the same set of committees and produce 14005614014756.",
"robustness_analysis": "Robustness 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 $\boxed{14005614014756}$.) |
math-001467 | Discrete Math: Choosing without Replacement | 1 | Work carefully and justify each inference: 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 11 people from 147 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections ... | [
{
"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: $\\boxed{11825811462719982}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{147}{11}$, so both methods count the same set of committees and produce 11825811462719982.",
"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... | 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-001468 | 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 4 people from 108 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: $nP k = 108\\cdot(107)\\cdots(105)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5359095}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{108}{4}$, so both methods count the same set of committees and produce 5359095.",
"robustness_analysis": "If the problem were perturbed: 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 $k!$'. |
math-001469 | Combinatorics: Committees — Combinations | 1 | Work carefully and justify each inference: 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 160 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2274048887320496}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{160}{10}$, so both methods count the same set of committees and produce 2274048887320496.",
"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 $k!$'. |
math-001470 | Combinatorics: Subsets — Binomial Coefficients | 1 | Provide a rigorous 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 187 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) ... | [
{
"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: $nP k = 187\\cdot(186)\\cdots(183)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1805568402}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{187}{5}$, so both methods count the same set of committees and produce 1805568402.",
"robustness_analysis": "Sensitivity... | [
{
"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 $\boxed{1805568402}$.) |
math-001471 | Combinatorics: Counting — Order vs Unordered | 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 3 people from 52 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: $nP k = 52\\cdot(51)\\cdots(50)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{22100}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{52}{3}$, so both methods count the same set of committees and produce 22100.",
"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... | 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 $\boxed{22100}$.) |
math-001472 | Combinatorics: Committees — Combinations | 1 | Where appropriate, name the theorem you use: 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 5 people from 71 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve... | [
{
"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: $nP k = 71\\cdot(70)\\cdots(67)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{13019909}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{71}{5}$, so both methods count the same set of committees and produce 13019909.",
"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 $k!$'. |
math-001473 | Discrete Math: Choosing without Replacement | 1 | Checkpoint: 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 144 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{767464189477128}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{144}{10}$, so both methods count the same set of committees and produce 767464189477128.",
"robustness_ana... | [
{
"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 $\boxed{767464189477128}$.) |
math-001474 | Discrete Math: Choosing without Replacement | 1 | Task: 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 15 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and div... | [
{
"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: $\\boxed{455}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{15}{3}$, so both methods count the same set of committees and produce 455.",
"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 $\boxed{455}$.) |
math-001475 | Discrete Math: Choosing without Replacement | 1 | Complete the analysis: 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 64 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: $nP k = 64\\cdot(63)\\cdots(57)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4426165368}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{64}{8}$, so both methods count the same set of committees and produce 4426165368.",
"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... | 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 $\boxed{4426165368}$.) |
math-001476 | Combinatorics: Counting Models — Subset Interpretation | 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 79 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: $nP k = 79\\cdot(78)\\cdots(75)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{22537515}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{79}{5}$, so both methods count the same set of committees and produce 22537515.",
"robustness_analysis": "Robustn... | [
{
"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 $\boxed{22537515}$.) |
math-001477 | Combinatorics: Counting — Order vs Unordered | 1 | Explain each transformation: 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 193 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting or... | [
{
"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: $\\boxed{66364016544}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{193}{6}$, so both methods count the same set of committees and produce 66364016544.",
"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... | 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 $\boxed{66364016544}$.) |
math-001478 | 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 7 people from 140 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered s... | [
{
"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: $nP k = 140\\cdot(139)\\cdots(134)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{179593009560}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{140}{7}$, so both methods count the same set of committees and produce 179593009560.",
"robustness_analysis": "Robustness note: 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 $k!$'. |
math-001479 | Combinatorics: Counting Models — Subset Interpretation | 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 8 people from 172 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{16104878212995}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{172}{8}$, so both methods count the same set of committees and produce 16104878212995.",
"robustness_analysis": "Rob... | [
{
"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-001480 | Combinatorics: Counting — Order vs Unordered | 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 9 people from 176 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: $\\boxed{362704129387600}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{176}{9}$, so both methods count the same set of committees and produce 362704129387600.",
"robustness_analysis": "I... | [
{
"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 $\boxed{362704129387600}$.) |
math-001481 | Discrete Math: Choosing without Replacement | 1 | Give reasoning, not just computation: 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 117 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and d... | [
{
"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: $nP k = 117\\cdot(116)\\cdots(108)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{89196105660948}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{117}{10}$, so both methods count the same set of committees and produce 89196105660948.",
"robustness_analysis": "Se... | [
{
"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 $\boxed{89196105660948}$.) |
math-001482 | Discrete Math: Choosing without Replacement | 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 2 people from 103 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5253}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{103}{2}$, so both methods count the same set of committees and produce 5253.",
"robustness_analysis": "Robustness 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 $\boxed{5253}$.) |
math-001483 | Combinatorics: Counting — Permute Then Divide | 1 | Track quantifiers carefully: 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 186 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{10677247751972451}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{186}{10}$, so both methods count the same set of committees and produce 10677247751972451.",
"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 $\boxed{10677247751972451}$.) |
math-001484 | Combinatorics: Counting Models — Subset Interpretation | 1 | Use two approaches if possible: 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 132 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{117850651776}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{132}{7}$, so both methods count the same set of committees and produce 117850651776.",
"robustness_analysis": "General... | [
{
"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 $\boxed{117850651776}$.) |
math-001485 | Combinatorics: Subsets — Binomial Coefficients | 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 5 people from 138 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: $nP k = 138\\cdot(137)\\cdots(134)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{387610812}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{138}{5}$, so both methods count the same set of committees and produce 387610812.",
"robustness_analysis": "Robustness note: 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... | 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 $\boxed{387610812}$.) |
math-001486 | Discrete Math: Choosing without Replacement | 1 | Answer using clear logical steps: 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 110 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by count... | [
{
"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: $nP k = 110\\cdot(109)\\cdots(101)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{46897636623981}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{110}{10}$, so both methods count the same set of committees and produce 46897636623981.",
"robustness_analysis": "Ro... | [
{
"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-001487 | Combinatorics: Counting Models — Subset Interpretation | 1 | Show all reasoning: 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 114 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one ... | [
{
"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: $nP k = 114\\cdot(113)\\cdots(111)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6672876}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{114}{4}$, so both methods count the same set of committees and produce 6672876.",
"robustness_analysis": "If the problem we... | [
{
"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 $\boxed{6672876}$.) |
math-001488 | 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 7 people from 78 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: $\\boxed{2641902120}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{78}{7}$, so both methods count the same set of committees and produce 2641902120.",
"robustness_analysis": "If the problem were perturbed: 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 $\boxed{2641902120}$.) |
math-001489 | Combinatorics: Counting — Order vs Unordered | 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 11 people from 160 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections a... | [
{
"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: $\\boxed{31009757554370400}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{160}{11}$, so both methods count the same set of committees and produce 31009757554370400.",
"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... | 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-001490 | Combinatorics: Counting — Order vs Unordered | 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 4 people from 192 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counti... | [
{
"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: $\\boxed{54870480}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{192}{4}$, so both methods count the same set of committees and produce 54870480.",
"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-001491 | Combinatorics: Committees — Combinations | 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 4 people from 158 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: $\\boxed{24992045}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{158}{4}$, so both methods count the same set of committees and produce 24992045.",
"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... | 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-001492 | Combinatorics: Subsets — Binomial Coefficients | 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 9 people from 99 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by countin... | [
{
"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: $\\boxed{1731030945644}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{99}{9}$, so both methods count the same set of committees and produce 1731030945644.",
"robustness_analysis": "Sensit... | [
{
"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 $\boxed{1731030945644}$.) |
math-001493 | Combinatorics: Committees — Combinations | 1 | Question: 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 143 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: $\\boxed{209004408899}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{143}{7}$, so both methods count the same set of committees and produce 209004408899.",
"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... | 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 $\boxed{209004408899}$.) |
math-001494 | 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 6 people from 184 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: $\\boxed{49637730324}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{184}{6}$, so both methods count the same set of committees and produce 49637730324.",
"robustness_analysis": "Robustnes... | [
{
"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 $\boxed{49637730324}$.) |
math-001495 | Combinatorics: Committees — Combinations | 1 | Explain what is being counted/optimized: 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 161 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: $\\boxed{26964280}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{161}{4}$, so both methods count the same set of committees and produce 26964280.",
"robustness_analysis": "Robust... | [
{
"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-001496 | Combinatorics: Counting — Permute Then Divide | 1 | Indicate where a theorem is used: 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 12 people from 49 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: $\\boxed{92263734836}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{49}{12}$, so both methods count the same set of committees and produce 92263734836.",
"robustness_analysis": "Robustness 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... | 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-001497 | Combinatorics: Subsets — Binomial Coefficients | 1 | Be explicit about assumptions: 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 152 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{632671880}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{152}{5}$, so both methods count the same set of committees and produce 632671880.",
"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 $\boxed{632671880}$.) |
math-001498 | 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 12 people from 32 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: $nP k = 32\\cdot(31)\\cdots(21)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{225792840}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{32}{12}$, so both methods count the same set of committees and produce 225792840.",
"robustness_analysis": "Generality note: 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 $k!$'. (Here the result is $\boxed{225792840}$.) |
math-001499 | Combinatorics: Committees — Combinations | 1 | Work carefully and justify each inference: 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 87 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve b... | [
{
"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: $nP k = 87\\cdot(86)\\cdots(86)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3741}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{87}{2}$, so both methods count the same set of committees and produce 3741.",
"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-001500 | Combinatorics: Counting — Permute Then Divide | 1 | Provide both a computational and a conceptual explanation: 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 139 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selecti... | [
{
"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: $\\boxed{170613359082}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{139}{7}$, so both methods count the same set of committees and produce 170613359082.",
"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... | 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 $\boxed{170613359082}$.) |
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