2020 AMC 8 第 10 题

先试着解答 2020 AMC 8 第 10 题,然后核对你的答案与视频与精心整理的解答,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 2020 AMC 8 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。

10.

扎拉有 44 个弹珠:阿吉、邦布尔比、斯蒂利和泰格。她想把它们排成一行展示在架子上,但不想把斯蒂利和泰格放在相邻位置。有多少种排法?
Zara has a collection of 44 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?
66
88
1212
1818
2424
答案:C
知识点:有限制的排列补集计数
小提示:
先数所有排列
Count all arrangements first
大提示:
减去把斯蒂利和泰格看成一个整体时的相邻排列
Subtract arrangements where Steelie and Tiger are treated as one block
视频讲解:
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文字解答:
四个弹珠共有 4!=244!=24 种排列。
若斯蒂利和泰格相邻,把它们看成一个整体,则有 3!3! 种放置整体和另外两个弹珠的方法,整体内部有 22 种顺序,共 3!⋅2=123!\cdot2=12 种。
不相邻的排法为 24−12=1224-12=12。
正确答案是 C。
There are 4!=244!=24 total arrangements of the marbles.
If the Steelie and Tiger are adjacent, treat them as one block. Then there are 3!3! ways to arrange the block with the other two marbles, and 22 orders inside the block, for 3!⋅2=123!\cdot2=12 adjacent arrangements.
Therefore, 24−12=1224-12=12 arrangements keep the Steelie and Tiger separated.
Thus, the correct answer is C.
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