AMC 10 · 2024 · #9
Grade 7 countingProblem
In how many ways can 6 juniors and 6 seniors form 3 disjoint teams of 4 people so
that each team has 2 juniors and 2 seniors?
Pick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: From $6$ juniors and $6$ seniors, count how many ways the $12$ students can be split into $3$ unlabeled teams of $4$ such that every team holds exactly $2$ juniors and $2$ seniors.
Givens: $6$ juniors and $6$ seniors, $12$ students in all; Form exactly $3$ teams, each with $4$ students; Each team must have $2$ juniors and $2$ seniors; Answer choices: (A) $720$, (B) $1350$, (C) $2700$, (D) $3280$, (E) $8100$
Unknowns: The number of ways to partition the students into the three teams
Understand
Restated: From $6$ juniors and $6$ seniors, count how many ways the $12$ students can be split into $3$ unlabeled teams of $4$ such that every team holds exactly $2$ juniors and $2$ seniors.
Givens: $6$ juniors and $6$ seniors, $12$ students in all; Form exactly $3$ teams, each with $4$ students; Each team must have $2$ juniors and $2$ seniors; Answer choices: (A) $720$, (B) $1350$, (C) $2700$, (D) $3280$, (E) $8100$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #16 Change Focus / Count the Complement
The setup mixes two independent choices (juniors, seniors) with three teams, so it screams for Tool #7 (Identify Subproblems). Pick the first team's $2$ juniors and $2$ seniors, then the second team's, then the third — each subproblem is a clean $\binom{n}{2}$ count, and multiplying gives the number of labeled lineups (Team A, Team B, Team C). Tool #16 (Change Focus) then handles the wording "$3$ disjoint teams": because the teams have no labels, every unordered partition has been counted $3!$ times. Dividing by $3!$ converts the labeled count into the unlabeled one — the answer the problem wants.
Execute — Answer: B
7.SP.C.8 Step 1 - Subproblem 1: build the first team (call it Team A for now).
- Choose $2$ of the $6$ juniors, then $2$ of the $6$ seniors.
- The two choices are independent, so multiply them.
💡 $\binom{6}{2}=15$ is just "pick $2$ from $6$". Pairing every junior choice with every senior choice is the Grade 7 fundamental counting principle.
7.SP.C.8 Step 2 - Subproblem 2: build the second team (Team B) from the remaining $4$ juniors and $4$ seniors.
- Same recipe, smaller pool.
💡 After taking $2$ from each group, $4$ juniors and $4$ seniors remain. The structure repeats — only the pool size shrinks.
7.SP.C.8 Step 3 - Subproblem 3: the third team (Team C) is forced.
- Only $2$ juniors and $2$ seniors are left, and they all must go together.
- There is exactly one way.
💡 When the pool size matches the choice size, $\binom{n}{n}=1$ — the leftover is the team.
7.SP.C.8 Step 4 Multiply the three subproblems to count the number of labeled lineups (Team A, Team B, Team C in order).
💡 Multiplication chains the three independent stages into one count of ordered team tuples.
7.SP.C.8 Step 5 - Change focus: the problem's teams are not labeled.
- Every unordered partition into three teams shows up in our $8100$ count once for each way to assign the labels A, B, C to the same three teams — that is $3! = 6$ times.
- Divide to remove the overcount.
💡 Switching from "ordered teams" to "unordered teams" is the classic Tool #16 move: count the easier (labeled) version, then divide by the number of relabelings ($3!$).
7.SP.C.8 Subproblem 1: build the first team (call it Team A for now). Choose $2$ of the $ 7.SP.C.8 Subproblem 2: build the second team (Team B) from the remaining $4$ juniors and 7.SP.C.8 Subproblem 3: the third team (Team C) is forced. Only $2$ juniors and $2$ senior 7.SP.C.8 Multiply the three subproblems to count the number of labeled lineups (Team A, T 7.SP.C.8 Change focus: the problem's teams are not labeled. Every unordered partition int Review
Reasonableness: Cross-check the $3!$ correction with a tiny case. Suppose just $2$ juniors $\{J_1,J_2\}$ and $2$ seniors $\{S_1,S_2\}$ are split into $1$ team of $4$. The labeled count is $\binom{2}{2}\binom{2}{2}=1$, and there is also only $1$ unordered team — matching $1/1!=1$. Now scale to $4$ juniors and $4$ seniors split into $2$ teams of $4$: labeled count $\binom{4}{2}\binom{4}{2}\cdot\binom{2}{2}\binom{2}{2}=36$, and unordered count is $36/2!=18$. Listing the $\binom{4}{2}=6$ ways to pick the junior pair for one team, paired with $6$ senior pairings, then halving for the label swap, gives the same $18$. The $/3!$ correction in the main problem follows the same pattern. Also: option (E) $8100$ is exactly the labeled count, which is the classic trap if a student forgets the relabeling — (B) $1350$ is six times smaller, as expected.
Alternative: Tool #9 (Solve an Easier Related Problem): replace the numbers with $2$ juniors, $2$ seniors, and $1$ team of $4$ — answer is $1$. Bump to $4$ juniors, $4$ seniors, and $2$ teams — labeled $36$, unordered $36/2!=18$. The pattern $\dfrac{\prod \binom{2k}{2}\binom{2k}{2}}{k!}$ generalizes, and with $k=3$ it yields $\dfrac{225\cdot 36\cdot 1}{6}=1350$. Same answer, reached by building intuition from small cases.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Counting the choices for each team as a compound event: independent junior and senior selections multiplied together, then chained across the three team-building stages.)7.NS.A.2Apply and extend previous understandings of multiplication and division to multiply and divide rational numbers (Combining the three stage counts by multiplication ($225 \times 36 \times 1 = 8100$) and dividing by $3! = 6$ to undo the label overcount.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Expressing and computing $\binom{n}{2}=\dfrac{n(n-1)}{2}$ and the factorial $3! = 3\times 2\times 1 = 6$ used in the final division.)
⭐ Pick the teams one at a time, then divide by $3!$ to forget the labels — that's how an AMC 10 counting problem collapses into Grade 7 "compound event" arithmetic.
⭐ Pick the teams one at a time, then divide by $3!$ to forget the labels — that's how an AMC 10 counting problem collapses into Grade 7 "compound event" arithmetic.
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