Question

六张卡片分别标有 $1$ 到 $6$ ，要排成一行。求满足如下条件的排列数：可以移走其中一张卡片，使剩下的五张卡片按升序或降序排列。

Six cards numbered $1$ through $6$ are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order.

答案

答案：52

知识点：排列有限制的排列分类讨论

难度评级：2390

解答：

先数移走某张卡后其余卡片为升序的排列。任何这样的排列都可以通过选择要移走的卡片（ $6$ 种方式），并把它插入到其余五张卡片按递增顺序写成的一行的 $6$ 个空位之一来得到，共有 $36$ 种构造。但完全有序的一行 $123456$ 会由所有 $6$ 种卡片选择得到，而由有序行中交换一对相邻卡片得到的 $5$ 个排列各会出现两次（把这一对中的任意一张越过另一张）。其余每种构造都给出不同的排列。

因此升序的数量为 $1 + 5 + (36 - 6 - 10) = 26$ ，由对称性，降序排列也有 $26$ 个。没有排列被两边同时计入：否则需要一个五张卡的升序子序列和一个五张卡的降序子序列，至少需要 $5 + 5 - 1 = 9$ 张卡。

总数为 $26 + 26 = 52$ 。

First count arrangements from which some card's removal leaves the rest ascending. Any such arrangement arises by choosing the card to remove ( $6$ ways) and inserting it into one of the $6$ gaps of the other five cards written in increasing order, for $36$ constructions. But the fully sorted row $123456$ arises from all $6$ card choices, and each of the $5$ arrangements obtained by swapping two adjacent cards of the sorted row arises twice (move either card of the pair past the other). Every other construction gives a distinct arrangement.

So the ascending count is $1 + 5 + (36 - 6 - 10) = 26,$ and by symmetry there are $26$ descending arrangements. No arrangement is counted in both totals: that would require an ascending and a descending subsequence of five cards, needing at least $5 + 5 - 1 = 9$ cards.

The total is $26 + 26 = 52.$

2020 AIME I 第 5 题

答案：52

难度评级：2390

解答：

其他年份的第 5 题