拍照排队

Picture Day

专题: Brainteaser / 脑筋急转弯
难度: L4
来源: OpenQuant

题目详情

十名身高各不相同的学生要排队拍照。摄影师要求最高的两名学生站在正中间两个位置，其余学生按离中心越远身高越严格递减的方式排列。问一共有多少种排法？

Ten students of distinct heights are lining up for a picture. The photographer requires that the two tallest students stand in the two center positions and that the remaining students line up such that the heights strictly descend outwards. How many ways can the students line up?

解析

把 10 个位置记为

1,2,3,4,5,6,7,8,9,10,

其中中间两个位置是 5 和 6。设学生按身高从高到低依次为 $h_1>h_2>\cdots>h_{10}$ 。

首先，最高的两名学生 $h_1,h_2$ 必须站在中间两个位置，因此这一步有 $2! = 2$ 种安排方式。

剩下的 8 名学生要分到左右两侧，而且每一侧都必须满足“越靠近中心越高，越往外越矮”。这意味着：

一旦选定哪 4 个人站左边，他们在左边的位置顺序就被唯一确定；
右边那 4 个人的顺序也同样被唯一确定。

所以关键只是在 8 名学生中选哪 4 名去左边，共有 $\binom{8}{4}=70$ 种方法。

总数为 $2!\times\binom{8}{4}=2\times70=\boxed{140}.$

Original Explanation

We can think of the positions as

1, 2, 3, 4, 5,6, 7, 8, 9, 10

with the two center positions $5$ and $6$ . Let the students be ordered by height $h_1 > h_2 > h_3 ... > h_{10}$ .

The two tallest $h_1, h_2$ must occupy positions 5 and 6. They can be arranged in $2!$ ways.
The remaining eight students must fill the four positions on the left (4, 3, 2, 1) and the four on the right (7, 8, 9, 10) so that heights strictly decrease as you move away from the center.

Once you choose which four of the eight go to the left side, their order along positions 4→1 is forced: the largest of the chosen four sits at position 4 (nearest the center), the next largest at 3, etc. The same holds for the four chosen for the right side.
Hence for each choice of the left-side subset there is exactly one valid ordering of all eight remaining students.

The number of ways to choose which four go to the left is $\begin{pmatrix}8 \\ 4\end{pmatrix}$

Multiply the choices:

2! \times \begin{pmatrix}8 \\ 4\end{pmatrix} = 2 * 70 = \boxed{140}