Problem

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?

Solution

The number of ways in which we can choose the juniors for the team are ${6\choose2}{4\choose2}{2\choose2}=90$ . Similarly, the number of ways to choose the seniors are the same, so the total is $90\cdot90=8100$ . But we must divide the number of permutations of three teams, which is $3!$ . Thus the answer is $\frac{8100}{3!}=\frac{8100}{6}=\boxed{1350}$ .

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2024 AMC 10A Problems/Problem 9

Problem

Solution