id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-001601 | Combinatorics: Counting — Order vs Unordered | 1 | Give a fully justified solution: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 11 people from 150 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividi... | [
{
"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: $\\boxed{14885236522829400}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{150}{11}$, so both methods count the same set of committees and produce 14885236522829400.",
"robustness... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{14885236522829400}$.) |
math-001602 | Combinatorics: Counting — Permute Then Divide | 1 | Derive the result step-by-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 10 people from 78 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Exp... | [
{
"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: $nP k = 78\\cdot(77)\\cdots(69)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1258315963905}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{78}{10}$, so both methods count the same set of committees and produce 1258315963905.",
"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... | 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 $\boxed{1258315963905}$.) |
math-001603 | Discrete Math: Choosing without Replacement | 1 | Problem: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 7 people from 94 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": "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: $\\boxed{10235867928}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{94}{7}$, so both methods count the same set of committees and produce 10235867928.",
"robustness_analysis": "Robustness note: Permute-th... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{10235867928}$.) |
math-001604 | Discrete Math: Choosing without Replacement | 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 6 people from 83 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and ... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{377447148}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{83}{6}$, so both methods count the same set of committees and produce 377447148.",
"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... | 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 $\boxed{377447148}$.) |
math-001605 | Combinatorics: Counting — Order vs Unordered | 1 | Answer using clear logical steps: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 4 people from 92 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: $nP k = 92\\cdot(91)\\cdots(89)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2794155}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{92}{4}$, so both methods count the same set of committees and produce 2794155.",
"robustness_analysis": "If the pr... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001606 | Combinatorics: Counting — Order vs Unordered | 1 | Solve with verification: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 7 people from 84 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: $nP k = 84\\cdot(83)\\cdots(78)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4529365776}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{84}{7}$, so both methods count the same set of committees and produce 4529365776.",
"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... | 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 $\boxed{4529365776}$.) |
math-001607 | Combinatorics: Counting Models — Subset Interpretation | 1 | Warm-up: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 9 people from 123 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sentence wh... | [
{
"method_name": "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: $\\boxed{13159493855365}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{123}{9}$, so both methods count the same set of committees and produce 13159493855365.",
"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 $k!$'. |
math-001608 | Combinatorics: Committees — Combinations | 1 | Derive the result step-by-step: 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 123 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividin... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1541096362225563}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{123}{11}$, so both methods count the same set of committees and produce 1541096362225563.",
"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... | 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-001609 | Combinatorics: Counting — Permute Then Divide | 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 5 people from 122 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explai... | [
{
"method_name": "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: $nP k = 122\\cdot(121)\\cdots(118)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{207288004}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{122}{5}$, so both methods count the same set of committees and produce 207288004.",
"robustness_analysis": "If t... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{207288004}$.) |
math-001610 | Combinatorics: Counting Models — Subset Interpretation | 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 2 people from 111 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: $nP k = 111\\cdot(110)\\cdots(110)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6105}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{111}{2}$, so both methods count the same set of committees and produce 6105.",
"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 $k!$'. (Here the result is $\boxed{6105}$.) |
math-001611 | Combinatorics: Subsets — Binomial Coefficients | 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 98 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing ... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{812325612855848}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{98}{12}$, so both methods count the same set of committees and produce 812325612855848.",
"robustness_anal... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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 $\boxed{812325612855848}$.) |
math-001612 | Combinatorics: Counting — Order vs Unordered | 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 5 people from 26 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: $nP k = 26\\cdot(25)\\cdots(22)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{65780}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{26}{5}$, so both methods count the same set of committees and produce 65780.",
"robustness_analysis": "If the proble... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{65780}$.) |
math-001613 | Combinatorics: Committees — Combinations | 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 8 people from 194 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": "Both approaches agree after simplification. Final answer: $\\boxed{42992032003272}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{194}{8}$, so both methods count the same set of committees and produce 42992032003272.",
"robustness_analysis": "If the problem were pertur... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{42992032003272}$.) |
math-001614 | 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 6 people from 96 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{927048304}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{96}{6}$, so both methods count the same set of committees and produce 927048304.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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 $\boxed{927048304}$.) |
math-001615 | Combinatorics: Counting — Permute Then Divide | 1 | Complete the analysis: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 9 people from 112 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: $nP k = 112\\cdot(111)\\cdots(104)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5494563394320}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{112}{9}$, so both methods count the same set of committees and produce 5494563394320.",
"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 $k!$'. (Here the result is $\boxed{5494563394320}$.) |
math-001616 | Combinatorics: Counting — Order vs Unordered | 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 9 people from 146 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: $\\boxed{64593041407180}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{146}{9}$, so both methods count the same set of committees and produce 64593041407180.",
"robustness_analysis": "Gen... | [
{
"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-001617 | Combinatorics: Counting — Permute Then Divide | 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 8 people from 148 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: $\\boxed{4709614623714}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{148}{8}$, so both methods count the same set of committees and produce 4709614623714.",
"robustness_analysis": "If th... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Key idea: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{4709614623714}$.) |
math-001618 | Combinatorics: Subsets — Binomial Coefficients | 1 | Start by stating any domain restrictions: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 5 people from 169 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: $nP k = 169\\cdot(168)\\cdots(165)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1082239158}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{169}{5}$, so both methods count the same set of committees and produce 1082239158.",
"robustness_analysis": "If the prob... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001619 | Combinatorics: Counting — Order vs Unordered | 1 | Proceed methodically: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 6 people from 45 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": "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: $\\boxed{8145060}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{45}{6}$, so both methods count the same set of committees and produce 8145060.",
"robustness_analysis": "Robustness note: Permute-then-divid... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001620 | Combinatorics: Counting Models — Subset Interpretation | 1 | Exercise: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 8 people from 168 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections an... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{13288305520413}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{168}{8}$, so both methods count the same set of committees and produce 13288305520413.",
"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... | 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-001621 | Combinatorics: Counting — Permute Then Divide | 1 | Solve with verification: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 6 people from 189 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{58429377468}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{189}{6}$, so both methods count the same set of committees and produce 58429377468.",
"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... | 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-001622 | Combinatorics: Subsets — Binomial Coefficients | 1 | Use two approaches if possible: 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 124 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividin... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 124\\cdot(123)\\cdots(115)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{163177723806526}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{124}{10}$, so both methods count the same set of committees and produce 163177723806526.",
"robustness_analysis": "If the problem were 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 $k!$'. |
math-001623 | Combinatorics: Counting — Permute Then Divide | 1 | Track units/moduli 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 8 people from 193 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting o... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 193\\cdot(192)\\cdots(186)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{41219164704168}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{193}{8}$, so both methods count the same set of committees and produce 41219164704168.",
"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... | 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 $\boxed{41219164704168}$.) |
math-001624 | Combinatorics: Counting — Permute Then Divide | 1 | Prompt: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 9 people from 133 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": "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: $\\boxed{27212042858000}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{133}{9}$, so both methods count the same set of committees and produce 27212042858000.",
"robustness_analysis": "If ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001625 | Combinatorics: Committees — Combinations | 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 4 people from 68 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing 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: $nP k = 68\\cdot(67)\\cdots(65)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{814385}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{68}{4}$, so both methods count the same set of committees and produce 814385.",
"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... | 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-001626 | Combinatorics: Committees — Combinations | 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 9 people from 169 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: $nP k = 169\\cdot(168)\\cdots(161)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{249524848105533}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{169}{9}$, so both methods count the same set of committees and produce 249524848105533.",
"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... | 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 $\boxed{249524848105533}$.) |
math-001627 | Combinatorics: Counting Models — Subset Interpretation | 1 | Give an answer and a quick verification: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 4 people from 176 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{38630900}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{176}{4}$, so both methods count the same set of committees and produce 38630900.",
"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... | 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 $\boxed{38630900}$.) |
math-001628 | Combinatorics: Counting — Permute Then Divide | 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 5 people from 72 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{13991544}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{72}{5}$, so both methods count the same set of committees and produce 13991544.",
"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 $k!$'. (Here the result is $\boxed{13991544}$.) |
math-001629 | Combinatorics: Counting — Permute Then Divide | 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 2 people from 144 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{10296}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{144}{2}$, so both methods count the same set of committees and produce 10296.",
"robustness_analysis": "Generality n... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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 $\boxed{10296}$.) |
math-001630 | Combinatorics: Committees — Combinations | 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 6 people from 106 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selecti... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 106\\cdot(105)\\cdots(101)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1705904746}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{106}{6}$, so both methods count the same set of committees and produce 1705904746.",
"robustness_analysis": "Robustness ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001631 | Combinatorics: Counting Models — Subset Interpretation | 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 5 people from 118 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{174963438}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{118}{5}$, so both methods count the same set of committees and produce 174963438.",
"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 $\boxed{174963438}$.) |
math-001632 | Combinatorics: Counting — Permute Then Divide | 1 | Explain why your operations are valid: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 6 people from 20 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: $nP k = 20\\cdot(19)\\cdots(15)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{38760}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{20}{6}$, so both methods count the same set of committees and produce 38760.",
"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 $k!$'. (Here the result is $\boxed{38760}$.) |
math-001633 | Combinatorics: Counting — Permute Then Divide | 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 9 people from 68 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{49280065120}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{68}{9}$, so both methods count the same set of committees and produce 49280065120.",
"robustness_analysis": "If the pro... | [
{
"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 $\boxed{49280065120}$.) |
math-001634 | Discrete Math: Choosing without Replacement | 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 7 people from 147 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c... | [
{
"method_name": "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: $nP k = 147\\cdot(146)\\cdots(141)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{254573763444}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{147}{7}$, so both methods count the same set of committees and produce 254573763444.",
"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 $\boxed{254573763444}$.) |
math-001635 | Combinatorics: Counting — Permute Then Divide | 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 6 people from 120 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3652745460}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{120}{6}$, so both methods count the same set of committees and produce 3652745460.",
"robustness_analysis": "Ge... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001636 | Discrete Math: Choosing without Replacement | 1 | Derive the result step-by-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 4 people from 58 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: $nP k = 58\\cdot(57)\\cdots(55)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{424270}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{58}{4}$, so both methods count the same set of committees and produce 424270.",
"robustness_analysis": "Generality ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001637 | Combinatorics: Counting — Permute Then Divide | 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 7 people from 155 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections an... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 155\\cdot(154)\\cdots(149)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{371716103550}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{155}{7}$, so both methods count the same set of committees and produce 371716103550.",
"robustness_analysis": "Generality 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 $\boxed{371716103550}$.) |
math-001638 | 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 4 people from 49 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: $nP k = 49\\cdot(48)\\cdots(46)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{211876}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{49}{4}$, so both methods count the same set of committees and produce 211876.",
"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 $\boxed{211876}$.) |
math-001639 | Combinatorics: Counting Models — Subset Interpretation | 1 | Problem: 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 89 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: $nP k = 89\\cdot(88)\\cdots(82)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{70625252863}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{89}{8}$, so both methods count the same set of committees and produce 70625252863.",
"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... | 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 $\boxed{70625252863}$.) |
math-001640 | Discrete Math: Choosing without Replacement | 1 | Explain each transformation: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 2 people from 100 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: $nP k = 100\\cdot(99)\\cdots(99)$.",
"Step 2: Each unordered committee corresponds ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4950}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{100}{2}$, so both methods count the same set of committees and produce 4950.",
"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 $k!$'. (Here the result is $\boxed{4950}$.) |
math-001641 | 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 9 people from 128 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 128\\cdot(127)\\cdots(120)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{19062702032000}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{128}{9}$, so both methods count the same set of committees and produce 19062702032000.",
"robustness_analys... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001642 | Combinatorics: Counting — Permute Then Divide | 1 | Prompt: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 6 people from 98 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: $nP k = 98\\cdot(97)\\cdots(93)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1052618392}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{98}{6}$, so both methods count the same set of committees and produce 1052618392.",
"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... | 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 $\boxed{1052618392}$.) |
math-001643 | Discrete Math: Choosing without Replacement | 1 | Solve and justify each step: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 12 people from 136 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting o... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 136\\cdot(135)\\cdots(125)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{50718300243156000}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{136}{12}$, so both methods count the same set of committees and produce 50718300243156000.",
"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-001644 | Combinatorics: Committees — Combinations | 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 5 people from 24 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{42504}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{24}{5}$, so both methods count the same set of committees and produce 42504.",
"robustness_analysis": "Sensitivity 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... | 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 $\boxed{42504}$.) |
math-001645 | 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 12 people from 26 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: $nP k = 26\\cdot(25)\\cdots(15)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{9657700}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{26}{12}$, so both methods count the same set of committees and produce 9657700.",
"robustness_analysis": "Sensitivity analysis: Permute-then-divid... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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 $\boxed{9657700}$.) |
math-001646 | Combinatorics: Counting Models — Subset Interpretation | 1 | Complete the analysis: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 11 people from 54 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: $nP k = 54\\cdot(53)\\cdots(44)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{95722852680}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{54}{11}$, so both methods count the same set of committees and produce 95722852680.",
"robustness_analysis": "If the problem were pertur... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{95722852680}$.) |
math-001647 | Combinatorics: Counting Models — Subset Interpretation | 1 | Solve and then verify: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 12 people from 134 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Expla... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 134\\cdot(133)\\cdots(123)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{42132653339249200}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{134}{12}$, so both methods count the same set of committees and produce 42132653339249200.",
"robustness... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{42132653339249200}$.) |
math-001648 | Combinatorics: Counting — Order vs Unordered | 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 12 people from 35 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in ... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{834451800}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{35}{12}$, so both methods count the same set of committees and produce 834451800.",
"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... | 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 $\boxed{834451800}$.) |
math-001649 | Discrete Math: Choosing without Replacement | 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 7 people from 171 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: $nP k = 171\\cdot(170)\\cdots(165)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{749064102930}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{171}{7}$, so both methods count the same set of committees and produce 749064102930.",
"robustness_analysis": "Sensitivity analysis: Pe... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{749064102930}$.) |
math-001650 | Combinatorics: Counting — Permute Then Divide | 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 4 people from 37 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{66045}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{37}{4}$, so both methods count the same set of committees and produce 66045.",
"robustness_analysis": "Robustness note: Permute-then-divide wo... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001651 | Combinatorics: Subsets — Binomial Coefficients | 1 | Give reasoning, not just computation: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 3 people from 98 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: $nP k = 98\\cdot(97)\\cdots(96)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{152096}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{98}{3}$, so both methods count the same set of committees and produce 152096.",
"robustness_analysis": "Robustness ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Takeaway: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{152096}$.) |
math-001652 | Discrete Math: Choosing without Replacement | 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 11 people from 38 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1203322288}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{38}{11}$, so both methods count the same set of committees and produce 1203322288.",
"robustness_analysis": "If... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001653 | Combinatorics: Counting — Permute Then Divide | 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 12 people from 165 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain i... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{564378143812462980}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{165}{12}$, so both methods count the same set of committees and produce 564378143812462980.",
"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... | 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 $\boxed{564378143812462980}$.) |
math-001654 | Discrete Math: Choosing without Replacement | 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 4 people from 69 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{864501}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{69}{4}$, so both methods count the same set of committees and produce 864501.",
"robustness_analysis": "Sensitivity analysis: Permute-then-di... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001655 | 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 9 people from 32 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{28048800}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{32}{9}$, so both methods count the same set of committees and produce 28048800.",
"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 $\boxed{28048800}$.) |
math-001656 | Combinatorics: Counting — Permute Then Divide | 1 | Give an answer and a quick verification: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 4 people from 48 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: $nP k = 48\\cdot(47)\\cdots(45)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{194580}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{48}{4}$, so both methods count the same set of committees and produce 194580.",
"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... | 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 $\boxed{194580}$.) |
math-001657 | Discrete Math: Choosing without Replacement | 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 2 people from 50 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1225}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{50}{2}$, so both methods count the same set of committees and produce 1225.",
"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... | 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 $\boxed{1225}$.) |
math-001658 | Combinatorics: Counting Models — Subset Interpretation | 1 | Explain why your operations are valid: 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 187 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1416167483302}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{187}{7}$, so both methods count the same set of committees and produce 1416167483302.",
"robustness_analysis": "Sensi... | [
{
"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 $\boxed{1416167483302}$.) |
math-001659 | Combinatorics: Committees — Combinations | 1 | Determine the requested value: 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 42 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 42\\cdot(41)\\cdots(32)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4280561376}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{42}{11}$, so both methods count the same set of committees and produce 4280561376.",
"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 $k!$'. |
math-001660 | Combinatorics: Counting — Permute Then Divide | 1 | Solve with verification: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 2 people from 123 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: $nP k = 123\\cdot(122)\\cdots(122)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7503}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{123}{2}$, so both methods count the same set of committees and produce 7503.",
"robustness_analysis": "Generality 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... | 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 $\boxed{7503}$.) |
math-001661 | Combinatorics: Subsets — Binomial Coefficients | 1 | Carefully track domains: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 12 people from 96 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: $\\boxed{624668654531480}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{96}{12}$, so both methods count the same set of committees and produce 624668654531480.",
"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... | 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 $\boxed{624668654531480}$.) |
math-001662 | Discrete Math: Choosing without Replacement | 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 11 people from 34 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: $nP k = 34\\cdot(33)\\cdots(24)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{286097760}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{34}{11}$, so both methods count the same set of committees and produce 286097760.",
"robustness_analysis": "If the problem were perturbed: 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 $\boxed{286097760}$.) |
math-001663 | Combinatorics: Counting — Permute Then Divide | 1 | Give a fully justified solution: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 10 people from 81 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: $\\boxed{1878392407320}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{81}{10}$, so both methods count the same set of committees and produce 1878392407320.",
"robustness_analysis": "Sensi... | [
{
"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-001664 | Combinatorics: Committees — Combinations | 1 | Work carefully and justify each inference: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 6 people from 199 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: $nP k = 199\\cdot(198)\\cdots(194)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{79936367511}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{199}{6}$, so both methods count the same set of committees and produce 79936367511.",
"robustness_analysis": "Generalit... | [
{
"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 $\boxed{79936367511}$.) |
math-001665 | Combinatorics: Subsets — Binomial Coefficients | 1 | Be explicit about assumptions: Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 2 people from 177 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: $\\boxed{15576}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{177}{2}$, so both methods count the same set of committees and produce 15576.",
"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 $k!$'. (Here the result is $\boxed{15576}$.) |
math-001666 | Combinatorics: Counting Models — Subset Interpretation | 1 | Determine the requested value: 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 138 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: $\\boxed{9453}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{138}{2}$, so both methods count the same set of committees and produce 9453.",
"robustness_analysis": "Robustness note: Permute-then-divide works whe... | [
{
"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-001667 | Combinatorics: Committees — Combinations | 1 | Make each step logically reversible (or explain if not): Count the number of unordered selections (no repetition). Provide both $\binom{n}{k}$ reasoning and an ordered-count cross-check:
How many ways are there to choose a committee of 6 people from 197 distinct people?
(a) Solve using the binomial coefficient definiti... | [
{
"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: $\\boxed{75176946208}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{197}{6}$, so both methods count the same set of committees and produce 75176946208.",
"robustness_analysis": "Generality 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... | 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-001668 | Combinatorics: Counting Models — Subset Interpretation | 1 | Try to avoid pattern-matching; explain why: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 12 people from 179 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: $nP k = 179\\cdot(178)\\cdots(168)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1549618771050715820}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{179}{12}$, so both methods count the same set of committees and produce 1549618771050715820.",
"robustness_analysis": "Generalit... | [
{
"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-001669 | Discrete Math: Choosing without Replacement | 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 12 people from 27 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: $nP k = 27\\cdot(26)\\cdots(16)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{17383860}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{27}{12}$, so both methods count the same set of committees and produce 17383860.",
"robustness_analysis": "If the problem were perturbed: P... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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 $\boxed{17383860}$.) |
math-001670 | Discrete Math: Choosing without Replacement | 1 | Give a fully justified solution: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 8 people from 165 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: $\\boxed{11468588169060}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{165}{8}$, so both methods count the same set of committees and produce 11468588169060.",
"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... | 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 $\boxed{11468588169060}$.) |
math-001671 | Combinatorics: Counting — Permute Then Divide | 1 | Solve and then verify: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 5 people from 183 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: $nP k = 183\\cdot(182)\\cdots(179)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1618621641}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{183}{5}$, so both methods count the same set of committees and produce 1618621641.",
"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 $k!$'. |
math-001672 | Combinatorics: Committees — Combinations | 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 8 people from 199 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) E... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{52895036330136}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{199}{8}$, so both methods count the same set of committees and produce 52895036330136.",
"robustness_analysis": "If ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{52895036330136}$.) |
math-001673 | Combinatorics: Counting — Permute Then Divide | 1 | Be explicit about assumptions: 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 25 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: $nP k = 25\\cdot(24)\\cdots(19)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{480700}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{25}{7}$, so both methods count the same set of committees and produce 480700.",
"robustness_analysis": "If the problem were ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001674 | Discrete Math: Choosing without Replacement | 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 11 people from 181 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by 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: $\\boxed{125519813934742980}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{181}{11}$, so both methods count the same set of committees and produce 125519813934742980.",
"robustness_analysis": "If the prob... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001675 | Combinatorics: Counting — Order vs Unordered | 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 7 people from 23 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: $\\boxed{245157}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{23}{7}$, so both methods count the same set of committees and produce 245157.",
"robustness_analysis": "Robustness note: Per... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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 $\boxed{245157}$.) |
math-001676 | Combinatorics: Subsets — Binomial Coefficients | 1 | Explain what is being counted/optimized: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 12 people from 183 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections an... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 183\\cdot(182)\\cdots(172)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2037311365940813690}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{183}{12}$, so both methods count the same set of committees and produce 2037311365940813690.",
"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... | 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 $\boxed{2037311365940813690}$.) |
math-001677 | Combinatorics: Counting — Order vs Unordered | 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 192 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selecti... | [
{
"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: $nP k = 192\\cdot(191)\\cdots(186)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1708566412608}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{192}{7}$, so both methods count the same set of committees and produce 1708566412608.",
"robustness_analysis": "Robustness note: Permute-the... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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 $\boxed{1708566412608}$.) |
math-001678 | Discrete Math: Choosing without Replacement | 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 3 people from 135 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: $\\boxed{400995}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{135}{3}$, so both methods count the same set of committees and produce 400995.",
"robustness_analysis": "If the problem were perturbed: 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 $k!$'. (Here the result is $\boxed{400995}$.) |
math-001679 | Combinatorics: Committees — Combinations | 1 | Give a fully justified solution: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 10 people from 58 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividin... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 58\\cdot(57)\\cdots(49)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{52179482355}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{58}{10}$, so both methods count the same set of committees and produce 52179482355.",
"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 $\boxed{52179482355}$.) |
math-001680 | Combinatorics: Counting — Permute Then Divide | 1 | Be explicit about assumptions: 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 141 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: $nP k = 141\\cdot(140)\\cdots(136)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{9798689908}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{141}{6}$, so both methods count the same set of committees and produce 9798689908.",
"robustness_analysis": "If the prob... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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 $\boxed{9798689908}$.) |
math-001681 | Combinatorics: Subsets — Binomial Coefficients | 1 | Problem: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 6 people from 172 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sentence wh... | [
{
"method_name": "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: $\\boxed{32927096748}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{172}{6}$, so both methods count the same set of committees and produce 32927096748.",
"robustness_analysis": "If the problem were perturbed: P... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{32927096748}$.) |
math-001682 | Combinatorics: Committees — Combinations | 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 12 people from 37 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: $\\boxed{1852482996}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{37}{12}$, so both methods count the same set of committees and produce 1852482996.",
"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 $k!$'. |
math-001683 | Combinatorics: Counting Models — Subset Interpretation | 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 10 people from 65 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: $nP k = 65\\cdot(64)\\cdots(56)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{179013799328}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{65}{10}$, so both methods count the same set of committees and produce 179013799328.",
"robustness_analysis": "If the problem were pert... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Remember: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. |
math-001684 | Combinatorics: Counting — Order vs Unordered | 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 6 people from 156 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": "Both approaches agree after simplification. Final answer: $\\boxed{18161699556}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{156}{6}$, so both methods count the same set of committees and produce 18161699556.",
"robustness_analysis": "Sensitivity analysis: Permute-th... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001685 | Combinatorics: Subsets — Binomial Coefficients | 1 | Proceed methodically: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 5 people from 170 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: $nP k = 170\\cdot(169)\\cdots(166)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1115034284}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{170}{5}$, so both methods count the same set of committees and produce 1115034284.",
"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 $k!$'. (Here the result is $\boxed{1115034284}$.) |
math-001686 | Combinatorics: Counting — Permute Then Divide | 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 11 people from 28 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: $\\boxed{21474180}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{28}{11}$, so both methods count the same set of committees and produce 21474180.",
"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-001687 | Combinatorics: Counting — Order vs Unordered | 1 | Give a theorem-based solution: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 9 people from 168 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{236236542585120}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{168}{9}$, so both methods count the same set of committees and produce 236236542585120.",
"robustness_analysis": "Sensitivity analys... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001688 | Combinatorics: Counting — Permute Then Divide | 1 | Solve (and briefly cross-validate): 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 110 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: $nP k = 110\\cdot(109)\\cdots(108)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{215820}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{110}{3}$, so both methods count the same set of committees and produce 215820.",
"robustness_analysis": "If the problem were perturbed: Permute-the... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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 $\boxed{215820}$.) |
math-001689 | Combinatorics: Counting Models — Subset Interpretation | 1 | Do not skip justification steps: How many committees are possible? Solve using two counting models (subset vs permute-and-divide):
How many ways are there to choose a committee of 4 people from 90 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: $nP k = 90\\cdot(89)\\cdots(87)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2555190}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{90}{4}$, so both methods count the same set of committees and produce 2555190.",
"robustness_analysis": "Sensitivi... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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-001690 | 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 3 people from 88 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: $nP k = 88\\cdot(87)\\cdots(86)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{109736}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{88}{3}$, so both methods count the same set of committees and produce 109736.",
"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 $k!$'. (Here the result is $\boxed{109736}$.) |
math-001691 | Combinatorics: Committees — Combinations | 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 4 people from 134 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{12840751}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{134}{4}$, so both methods count the same set of committees and produce 12840751.",
"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... | 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 $\boxed{12840751}$.) |
math-001692 | 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 4 people from 33 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sent... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 33\\cdot(32)\\cdots(30)$.",
"Step 2: Each unordered committee corresponds t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{40920}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{33}{4}$, so both methods count the same set of committees and produce 40920.",
"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... | 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 $\boxed{40920}$.) |
math-001693 | Combinatorics: Committees — Combinations | 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 8 people from 176 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": "Both approaches agree after simplification. Final answer: $\\boxed{19430578360050}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{176}{8}$, so both methods count the same set of committees and produce 19430578360050.",
"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... | 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 $\boxed{19430578360050}$.) |
math-001694 | Combinatorics: Counting — Permute Then Divide | 1 | Exercise: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 11 people from 92 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in one sent... | [
{
"method_name": "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: $\\boxed{53752237906276}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{92}{11}$, so both methods count the same set of committees and produce 53752237906276.",
"robustness_analys... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | 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 $\boxed{53752237906276}$.) |
math-001695 | Combinatorics: Subsets — Binomial Coefficients | 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 11 people from 124 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in... | [
{
"method_name": "Direct Binomial Coefficient",
"approach": "A committee is an unordered $k$-subset of an $n$-set, counted by $\\binom{n}{k}$.",
"steps": [
"Step 1: Order does not matter and no one can be selected twice, so we count $k$-element subsets of an $n$-element set.",
"Step 2: By de... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1691114592176724}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{124}{11}$, so both methods count the same set of committees and produce 1691114592176724.",
"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 $k!$'. |
math-001696 | Combinatorics: Counting Models — Subset Interpretation | 1 | Do not skip justification 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 5 people from 104 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by countin... | [
{
"method_name": "Permute-Then-Divide",
"approach": "Count ordered selections ($nP k$) and quotient by the number of internal permutations ($k!$) per committee.",
"steps": [
"Step 1: Count ordered selections: $nP k = 104\\cdot(103)\\cdots(100)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{91962520}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{104}{5}$, so both methods count the same set of committees and produce 91962520.",
"robustness_analysis": "Genera... | [
{
"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 $\boxed{91962520}$.) |
math-001697 | Combinatorics: Counting — Permute Then Divide | 1 | Proceed methodically: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 7 people from 184 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Explain in 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: $nP k = 184\\cdot(183)\\cdots(178)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1262216571096}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{184}{7}$, so both methods count the same set of committees and produce 1262216571096.",
"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 $k!$'. |
math-001698 | Discrete Math: Choosing without Replacement | 1 | Give a theorem-based solution: Give the exact count and a short explanation of why dividing by $k!$ is correct:
How many ways are there to choose a committee of 12 people from 185 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and dividing by $k!$.
(c) Exp... | [
{
"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: $nP k = 185\\cdot(184)\\cdots(174)$.",
"Step 2: Each unordered committee correspond... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2330625046935922100}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{185}{12}$, so both methods count the same set of committees and produce 2330625046935922100.",
"robustness_analysis": "If the problem ... | [
{
"error_description": "Used $n^k$ (allowing repetition and order) for committee selection.",
"why_plausible": "Counting functions from $\\{1,\\dots,k\\}$ to $\\{1,\\dots,n\\}$ is a common reflex.",
"why_wrong": "Committees require distinct elements and ignore order; $n^k$ counts ordered selections with... | Core principle: When order doesn’t matter and repetition is forbidden, count $k$-subsets: $\binom{n}{k}$. A reliable cross-check is 'ordered count divided by $k!$'. (Here the result is $\boxed{2330625046935922100}$.) |
math-001699 | Combinatorics: Subsets — Binomial Coefficients | 1 | Give reasoning, not just computation: Compute the number of ways to choose the group, and justify why order does not matter:
How many ways are there to choose a committee of 3 people from 154 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{596904}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{154}{3}$, so both methods count the same set of committees and produce 596904.",
"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... | 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-001700 | Combinatorics: Subsets — Binomial Coefficients | 1 | Solve (and briefly cross-validate): 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 93 distinct people?
(a) Solve using the binomial coefficient definition.
(b) Solve by counting ordered selections and divi... | [
{
"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: $\\boxed{8079421007658}$.\nThe permute-then-divide computation simplifies exactly to $\\binom{93}{10}$, so both methods count the same set of committees and produce 8079421007658.",
"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... | 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 $\boxed{8079421007658}$.) |
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