Question

来自三个不同家庭的三对年幼兄妹需要乘面包车出行。这六个孩子将坐在车的第二排和第三排，每排有三个座位。为了避免打闹，兄妹不能在同一排相邻而坐，也不能一个正好坐在另一个的正前方。共有多少种座位安排？

Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?

$60$

$72$

$92$

$96$

$120$

答案：D

知识点：分类讨论排列错位排列

难度评级：1930

解答：

若某个家庭的两个孩子坐在同一排，他们只能坐在该排的第 $1$ 和第 $3$ 个座位，这会迫使另一个家庭的两个孩子坐在同一列，违反条件。因此每排必须恰好有每个家庭的一个孩子。第二排安排三个家庭有 $3! = 6$ 种。第三排必须在每一列都与第二排家庭不同，是三个元素的错排，共 $2$ 种。最后每对兄妹可互换座位，共 $2^3 = 8$ 种。因此总数为 $6 \cdot 2 \cdot 8 = 96$ 。正确答案是 D。

Suppose some family put both children in one row. They'd have to take the non-adjacent seats $1$ and $3,$ which forces the middle family's two children into the same column. Not allowed. So each row holds exactly one child from each family. The second row is a permutation of the three families, $3! = 6$ ways. The third row needs a different family in every column, a derangement of the second row's order, and there are $2$ of those. Finally, each pair can swap its two children between their seats, $2^3 = 8$ ways. The total is $6 \cdot 2 \cdot 8 = 96.$ Therefore, the answer is D.

2018 AMC 10B 第 18 题

答案：D

难度评级：1930

解答：

其他年份的第 18 题