Zara有4个玻璃球，分别叫Aggie，Bumblebee，Steelie和Tiger。她想把它们在架子上排成一行，但不想让Steelie和Tiger相邻。那么她有多少种排列方法？

A. 6

B. 8

C. 12

D. 18

E. 24

1. 2020 AMC 8 官方解析

There are 6 ways Zara can place the Steelie and the Tiger so that they are not next to one another. Those arrangements are shown below.

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For each arrangement, there are 2 ways to place the remaining marbles: either the Aggie appears to the left of the Bumblebee or it appears to the right. This makes $6 \cdot 2 = 12$ arrangements with the Steelie and the Tiger not next to one another.

Zara 可以使用6种不同的方式放置 Steelie 和Tiger，使它们彼此不相邻，如下图所示

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对于每种排列方式，都有2种不同的方式放置剩余的弹珠：Aggie 出现在 Bumblebee 的左边或右边．因此有 $6 \cdot 2 = 12$ 种排法使得 Steelie 和 Tiger 不相邻．

The number of arrangements can be determined by using complementary counting. That is, begin with the total number of possible arrangements, which is $4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$, then subtract the number of ways to place the Steelie and the Tiger next to each other. If the Steelie and the Tiger are considered as a single object, then there are 3 objects to arrange and $3! = 3 \cdot 2 \cdot 1 = 6$ ways to do so. There are 2 ways to arrange the Steelie and the Tiger, so there are $6 \cdot 2 = 12$ arrangements with the Steelie and the Tiger adjacent to each other. Thus there are $24 - 12 = 12$ arrangements with the Steelie and the Tiger not next to one another.

可以通过使用补充计数来确定布置的数量．先计算所有可能的排列方式的总数，即 $4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$，然后减去使得 Steelie 和 Tiger 彼此相邻的情况．如果将 Steelie 和 Tiger 看成一个整体，则可以把四颗弹珠看成3个对象，并使用 $3! = 3 \cdot 2 \cdot 1 = 6$ 的方式进行排列．且在Steelie 和 Tiger 组成的对象中有2种方式来放置 Steelie 和 Tiger，故总共有 $6 \cdot 2 = 12$ 种排列方式使得 Steelie 和 Tiger 彼此相邻．因此，有 $24 - 12 = 12$ 种排法使得 Steelie 和 Tiger 不相邻．