2024 AMC 10A 第 9 题

Question

有

6

名十一年级生和

6

名十二年级生，要组成

3

个互不重叠的

4

人队伍，每队有

2

名十一年级生和

2

名十二年级生。有多少种方法？

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

将十一年级生分成三个无序对，将十二年级生也分成三个无序对。

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

每组学生各有

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

种分法；再用

3!

种方式匹配十一年级生对子和十二年级生对子。

There are

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

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

3!

ways

将

6

名十一年级生分成三个无序对，有

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

种方法。十二年级生同样有

15

种分法。每个队伍由一个十一年级生对子和一个十二年级生对子组成，所以还要将三个十一年级生对子与三个十二年级生对子配对，有

3! = 6

种方式。总数为

15 \cdot 15 \cdot 6 = 1350

，正确答案是 B。

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.

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