AMC8 2026

AMC8 2026 · Q17

AMC8 2026 · Q17. It mainly tests Permutations, Casework.

Four students are seated in a row. They chat with the people sitting next to them, then rearrange themselves so that they are no longer seated next to any of the same people. How many rearrangements are possible?

四个学生排成一排坐着。他们与相邻的人聊天，然后重新排列自己，使得他们不再与任何相同的人相邻。可能的重新排列数量有多少？

(A) 2 2

(B) 4 4

(D) 12 12

(E) 24 24

Answer

Correct choice: (A)

正确答案：（A）

Solution

Label the human beings A, B, C, and D in that order from left to right. It doesn't matter the original configuration. Now, the problem is: How many four letter strings ABCD exist such that A is not next to B, B not next to C, and C is not next to D. We can count the number of strings manually (or just do casework), to obtain $\boxed{2}$. This is incredibly similar to one AMC 10 problem, that asked how many four letter strings abcd have no two consecutive letters in the alphabet next to one another.

将四个人从左到右依次标记为 A、B、C 和 D。原始排列没有关系。现在问题是：有多少个四字母字符串 ABCD 存在，使得 A 不与 B 相邻，B 不与 C 相邻，且 C 不与 D 相邻。我们可以通过手动数数（或者分类讨论）来计算，得到答案为 $\boxed{2}$。这个问题与一个 AMC 10 题非常相似，那个题目问有多少四字母字符串 abcd 中没有任何两个字母在字母表中相邻。

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