Question

Zara 有 $4$ 个弹珠：Aggie、Bumblebee、Steelie 和 Tiger。她想把它们排成一行展示在架子上，但不想把 Steelie 和 Tiger 放在相邻位置。有多少种排法？

Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?

$6$

$8$

$12$

$18$

$24$

答案：C

知识点：有限制的排列补集计数

难度评级：960

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文字解答：

四个弹珠共有 $4!=24$ 种排列。

若 Steelie 和 Tiger 相邻，把它们看成一个整体，则有 $3!$ 种放置整体和另外两个弹珠的方法，整体内部有 $2$ 种顺序，共 $3!\cdot2=12$ 种。

不相邻的排法为 $24-12=12$ 。

正确答案是 C。

There are $4!=24$ total arrangements of the marbles.

If the Steelie and Tiger are adjacent, treat them as one block. Then there are $3!$ ways to arrange the block with the other two marbles, and $2$ orders inside the block, for $3!\cdot2=12$ adjacent arrangements.

Therefore, $24-12=12$ arrangements keep the Steelie and Tiger separated.

Thus, the correct answer is C.

2020 AMC 8 第 10 题

答案：C

难度评级：960

视频讲解：

文字解答：

其他年份的第 10 题