AMC 10 · 2024 · #9

Grade 7 counting

Archetype Casework by Position / Vertex Type └ Sub-archetype Arrangement Block Type

combinations-basicsystematic-enumerationset-partition identify-subproblemscasework ↑ Prerequisites: factorialmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

Problem

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.

Pick an answer.

(A)

720

(B)

1350

(C)

2700

(D)

3280

(E)

8100

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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 C(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.

1STEP 1

Subproblem 1 (Team A): pick 2 of 6 juniors and 2 of 6 seniors — independent choices, so multiply for 225 ways.

C(6, 2) × C(6, 2) = 15 × 15 = 225

2STEP 2

Subproblem 2 (Team B): same recipe on the leftover 4 juniors and 4 seniors — 36 ways.

C(4, 2) × C(4, 2) = 6 × 6 = 36

3STEP 3

Subproblem 3 (Team C): only 2 juniors and 2 seniors remain — forced, exactly 1 way.

C(2, 2) × C(2, 2) = 1 × 1 = 1

4STEP 4

Multiply the three stages for the labeled lineups (Team A, B, C in order): 225 × 36 × 1 = 8100.

225 × 36 × 1 = 8100 labeled lineups

5STEP 5

Teams are unlabeled, so each partition was counted 3! = 6 times; divide: 8100 ÷ 6 = 1350 → (B).

\frac{8100}{3!}

\frac{8100}{6}

= 1350 → (B)

Answer

1350

Cross-check the 3! correction with a tiny case. Suppose just 2 juniors {J₁,J₂} and 2 seniors {S₁,S₂} are split into 1 team of 4. The labeled count is C(2, 2)C(2, 2)=1, and there is also only 1 unordered team — matching

\frac{1}{1!}

=1. Now scale to 4 juniors and 4 seniors split into 2 teams of 4: labeled count C(4, 2)C(4, 2)·C(2, 2)C(2, 2)=36, and unordered count is

\frac{36}{2!}

=18. Listing the C(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.

💡Key takeaway

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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