1994 AMC 12 第 22 题

先试着解答 1994 AMC 12 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 1994 AMC 12 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.

一排九把椅子将由六名学生和阿尔法、贝塔、伽马三位教授就座。三位教授比六名学生先到,并决定选择座位,使每位教授的两边都是学生。阿尔法、贝塔和伽马三位教授有多少种选座方式?
Nine chairs in a row are to be occupied by six students and Professors Alpha, Beta and Gamma. These three professors arrive before the six students and decide to choose their chairs so that each professor will be between two students. In how many ways can Professors Alpha, Beta and Gamma choose their chairs?
1212
3636
6060
8484
630630
答案:C
知识点:seating arrangementsnonconsecutive selections
小提示:
教授的座位不能在两端,也不能彼此相邻
Professor chairs cannot be endpoints or adjacent to one another
大提示:
先从第 22 把至第 88 把椅子中选出三个互不相邻的位置,再安排三位教授
First choose three nonconsecutive positions from chairs 22 through 88, then assign the professors
解答:
教授必须占据第 2,3,…,82,3,\ldots,8 把椅子中的三个互不相邻的位置。这种 33 元子集从连续 77 个位置中选取,其数量为 (7−3+13)=(53)=10。 \binom{7-3+1}{3}=\binom53=10\text{。}三位不同的教授可以用 3!=63!=6 种方式安排到所选椅子上,共有 10⋅6=6010\cdot6=60 种。因此正确答案是 C。
The professors must occupy three nonconsecutive positions among chairs 2,3,…,8.2,3,\ldots,8. The number of such 33-subsets of 77 consecutive positions is (7−3+13)=(53)=10. \binom{7-3+1}{3}=\binom53=10. The three distinct professors can be assigned to the selected chairs in 3!=63!=6 ways, for 10⋅6=60.10\cdot6=60. Thus the correct answer is C.
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