Question

六把椅子均匀地围绕一张圆桌摆放。每把椅子上坐着一个人。每个人起身后坐到一把不是原来的椅子、也不与原来椅子相邻的椅子上，并且最后仍然每把椅子坐一个人。这样的方式有多少种？

Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same chair and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?

$14$

$16$

$18$

$20$

$24$

答案：D

知识点：环形排列分类讨论

难度评级：2310

小提示：

重新标号，使每个人必须留在原位或移到相邻椅子

Relabel so that each person must stay put or move to an adjacent chair

大提示：

保持其（重新标号后的）座位不变的人数必须是偶数： $0$ ， $2$ ， $4$ ，或 $6$ 。

The number of people who keep their (relabeled) seat must be even: $0,$ $2,$ $4,$ or $6$

解答：

先想象每个人都移动到正对面的椅子。条件变为：每个人必须坐在同一把椅子或相邻椅子上。保持原座位的人数必须为偶数（否则一个奇数长度的空段无法填满）。

若 $0$ 人保持原座位，所有人一齐向左移、一齐向右移，或与相邻者两两交换：共 $4$ 种。若 $2$ 人保持原座位，这两人必须相对或相邻，给出 $3+6=9$ 种选择，其余的人只能与相邻者两两交换。若 $4$ 人保持原座位，另外两人必须占据相邻座位并互相交换，给出 $6$ 种选择。若全部 $6$ 人都不动，有 $1$ 种。总数是 $4+9+6+1=20$ 。

因此，正确选项是 D。

First imagine everyone moves to the chair directly opposite. The condition becomes: each person must sit in the same chair or an adjacent one. The number of people who keep their seat must be even (otherwise an odd-length gap cannot be filled).

If $0$ keep their seat, everyone shifts left, shifts right, or swaps with a neighbor: $4$ ways. If $2$ keep their seats, those two must be opposite or adjacent, giving $3+6=9$ choices, and the remaining people are forced to swap in adjacent pairs. If $4$ keep their seats, the other two must occupy adjacent seats and swap, giving $6$ choices. If all $6$ stay, there is $1$ way. The total is $4+9+6+1=20.$

Thus, the correct answer is D.

2015 AMC 12B 第 22 题

答案：D

难度评级：2310

小提示：

大提示：

解答：

其他年份的第 22 题