2024 AMC 10A Problem 9

Question

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?

720

1350

2700

3280

8100

Split the juniors into three unordered pairs and the seniors into three unordered pairs

There are

\frac{6!}{2^3 \cdot 3!} = 15

such splits for each group; then match junior-pairs to senior-pairs in

3!

ways

Split the

6

juniors into three unordered pairs. There are

\frac{6!}{2!^3 3!} = 15

ways, and the same

15

for the seniors. Each team is one junior-pair paired with one senior-pair, so we match the three junior-pairs to the three senior-pairs in

3! = 6

ways. That’s

15 \cdot 15 \cdot 6 = 1350

sets of teams. Thus, B is the correct answer.

Problem 9 in other years