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?

$720$

$1350$

$2700$

$3280$

$8100$

答案：B

知识点：组合乘法原理

难度评级：1350

解答：

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.

2024 AMC 10A 第 9 题

答案：B

难度评级：1350

解答：

其他年份的第 9 题