Question

有 $13$ 张编号为 $1, 2, 3, \cdots, 13$ 的卡片排成一行。任务是按数字从小到大的顺序拿起这些卡片，并且反复从左到右扫描。在下面例子中，第一次扫描拿起 $1, 2, 3$ ，第二次拿起 $4$ , $5$ ，第三次拿起 $6$ ，第四次拿起 $7, 8, 9, 10$ ，第五次拿起 $11, 12, 13$ 。在所有 $13!$ 种排列中，有多少种排列会使 $13$ 张卡片恰好用两次扫描拿完？

Suppose that $13$ cards numbered $1, 2, 3, \cdots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?

$4082$

$4095$

$4096$

$8178$

$8191$

答案：D

知识点：组合补集计数

难度评级：1660

解答：

设第一次扫描拿到 $n$ 张卡片，其中 $1 \leq n \leq 12$ 。

选择这 $n$ 张卡片所占的位置后，剩余卡片的位置也被确定，因为两组卡片都必须按递增顺序排列。

共有 $\binom{13}{n}$ 种方法选择这 $n$ 张卡片的位置；但若这 $n$ 张卡片恰好位于最前面，则所有卡片都会在第一次扫描中拿完。

因此对固定的 $n$ ，共有 $\binom{13}{n} - 1$ 种排列。

对 $1\le n\le12$ 求和，得到 $\begin{gathered} \sum_{n=1}^{12}\left(\binom{13}{n}-1\right) \\ = (2^{13}-2)-12 \\ = 8178. \end{gathered}$

所以正确答案是 D。

Let $n$ be the number of cards picked up on the first pass, where $1 \leq n \leq 12.$

If we choose the spaces that the $n$ cards occupy, the positions of the remaining cards are determined since they must be placed in order.

There are $\binom{13}{n}$ ways to choose where the $n$ cards go, but if the $n$ cards are placed at the very beginning, then all the cards will be picked up on the first pass.

Therefore, for a given $n$ there are $\binom{13}{n} - 1$ ways to arrange the cards.

Summing over $1\le n\le12,$ we get $\begin{gathered} \sum_{n=1}^{12}\left(\binom{13}{n}-1\right) \\ = (2^{13}-2)-12 \\ = 8178. \end{gathered}$

Thus, D is the correct answer.

2022 AMC 10A 第 22 题

答案：D

难度评级：1660

解答：

其他年份的第 22 题