2024 AMC 10A Problem 9

P1-10~3 minPrint

Basic Counting Distributions & Stars and Bars Set up the equation/formula and compute; no special trick needed Exploit symmetry to reduce work or pair up objects

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?

Show solution without answering

Key insight

Split the juniors into three pairs (15 ways), split the seniors into three pairs (15 ways), then match each junior pair with a senior pair (6 ways).

Solution

Each team is $2$ juniors plus $2$ seniors. So build the teams in three stages: pair up the juniors, pair up the seniors, then decide which junior pair goes with which senior pair.

1. Pair up the $6$ juniors. Look at the first junior on the list. She can be paired with any of the other $5$ . Of the $4$ juniors left, look at the first one; he can be paired with any of the other $3$ . The last $2$ form the last pair. So there are $5 \cdot 3 \cdot 1 = 15$ ways.
(Picking the first junior's partner this way means each set of three pairs is counted exactly once, so there is nothing to divide by.)
2. Pair up the $6$ seniors. Exactly the same count: $15$ ways.
3. Match the pairs. Now there are $3$ junior pairs and $3$ senior pairs. The first junior pair can join any of the $3$ senior pairs, the second junior pair any of the $2$ left, and the last junior pair gets the last one. That is $3 \cdot 2 \cdot 1 = 6$ ways.
4. Multiply the three stages: $15 \cdot 15 \cdot 6 = 1350$ .

The answer is $\boxed{\textbf{(B)}\ 1350}$ .

Why this works

The teams have no names, so a count that picks "the first team, then the second team" counts each grouping several times and must be divided. Building the teams in stages (pair, pair, match) gives small counts that are easy to trust. The trick in step 1, always pairing the first person left on the list, is a handy way to count pairings without any dividing.

Alternative approach

Pretend the teams have names: Team 1, Team 2, Team 3.

Team 1 chooses $2$ of the $6$ juniors ( $15$ ways) and $2$ of the $6$ seniors ( $15$ ways): $225$ ways.
Team 2 chooses $2$ of the $4$ juniors left ( $6$ ways) and $2$ of the $4$ seniors left ( $6$ ways): $36$ ways.
Team 3 gets whoever is left: $1$ way.

That gives $225 \cdot 36 = 8100$ . But the real teams have no names, and each grouping was counted once for every way to hand out the three names, which is $3 \cdot 2 \cdot 1 = 6$ ways. So the answer is $8100 \div 6 = 1350$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

First pair up the juniors. Choosing pairs in order gives $\binom{6}{2}\binom{4}{2}\binom{2}{2} = 15\cdot 6\cdot 1 = 90$ , but the three pairs are not ordered, so divide by $3!$ : there are $90/6 = 15$ ways to split the juniors into three pairs. By the same count there are $15$ ways to split the seniors into three pairs.

Now each junior pair must be joined with one senior pair. Matching three junior pairs to three senior pairs is a permutation: $3! = 6$ ways.

Total: $15 \cdot 15 \cdot 6 = 1350$ .

The answer is $\boxed{\textbf{(B)}\ 1350}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Intermediate Counting & Probability, Chapter 7: Distributions (§7.2, §7.3)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques