2024 AMC 10A Problem 9

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 $\boxed{\textbf{(B)}\ 1350}$.

Teams are unordered, so whenever you pick them one at a time you must divide by the number of orders. Building the answer as "pair the juniors, pair the seniors, then match" makes every factor a small familiar count and forces you to notice that the matching step is a bijection between two sets of three pairs.

Count with labeled teams, then unlabel: Team 1 takes $\binom{6}{2}\binom{6}{2} = 225$ choices, Team 2 takes $\binom{4}{2}\binom{4}{2} = 36$, Team 3 takes the rest; that is $225\cdot 36 = 8100$ labeled outcomes, and dividing by $3!$ gives $1350$.

• Treating the three teams as labeled, which overcounts by 3! and gives 8100; or omitting the 3! matching factor and getting 225.
• Dividing by $3!$ twice (once for juniors and once for seniors) after already matching the pairs.