2015 AMC 12B 第 22 题

先试着解答 2015 AMC 12B 第 22 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2015 AMC 12B 解答,或核对答案

所有题目均经美国数学协会(MAA)官方合法授权使用。

22.

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?

1414

1616

1818

2020

2424

答案:D
知识点:环形排列分类讨论
难度评级:2310
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)。

00 人保持原座位,所有人向左移、向右移,或与相邻者交换:共 44 种。若 22 人保持原座位,这两人相对或相邻,给出 3+6=93+6=9 种,其余被迫确定。若 44 人保持原座位,有 66 种方法选择他们,另外两人交换。若全部 66 人都不动,有 11 种。总数是 4+9+6+1=204+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 00 keep their seat, everyone shifts left, shifts right, or swaps with a neighbor: 44 ways. If 22 keep their seats, those two must be opposite or adjacent, giving 3+6=93+6=9 choices, and the remaining people are forced to swap in adjacent pairs. If 44 keep their seats, the other two must occupy adjacent seats and swap, giving 66 choices. If all 66 stay, there is 11 way. The total is 4+9+6+1=20.4+9+6+1=20.

Thus, the correct answer is D.

← 第 21 题#21
完整试卷

其他年份的第 22 题