AMC10 2022 A

AMC10 2022 A · Q22

AMC10 2022 A · Q22. It mainly tests Basic counting (rules of product/sum), Combinations.

Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ 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?

假设有13张编号为$1, 2, 3, \ldots, 13$的卡片排成一排。任务是按数字递增顺序从左到右反复捡拾它们。在下面的例子中，第一遍捡起卡片$1, 2, 3$，第二遍捡起$4$和$5$，第三遍捡起$6$，第四遍捡起$7, 8, 9, 10$，第五遍捡起$11, 12, 13$。在13!种可能的卡片排列中，有多少种排列会在恰好两次捡拾中捡起所有13张卡片？

(A) 4082 4082

(B) 4095 4095

(D) 8178 8178

(E) 8191 8191

Answer

Correct choice: (D)

正确答案：（D）

Solution

For $1\leq k\leq 12,$ suppose that cards $1, 2, \ldots, k$ are picked up on the first pass. It follows that cards $k+1,k+2,\ldots,13$ are picked up on the second pass. Once we pick the spots for the cards on the first pass, there is only one way to arrange all $\boldsymbol{13}$ cards. For each value of $k,$ there are $\binom{13}{k}-1$ ways to pick the $k$ spots for the cards on the first pass: We exclude the arrangement in which the cards are arranged such that the first pass consists of all $13$ cards. Therefore, the answer is \[\sum_{k=1}^{12}\left[\binom{13}{k}-1\right] = \left[\sum_{k=1}^{12}\binom{13}{k}\right]-12 = \left[\sum_{k=0}^{13}\binom{13}{k}\right]-14 = 2^{13} - 14 = \boxed{\textbf{(D) } 8178}.\]

对于$1\leq k\leq 12$，假设第一遍捡起卡片$1, 2, \ldots, k$。由此，第一遍捡起卡片后，卡片$k+1,k+2,\ldots,13$将在第二遍被捡起。一旦我们为第一遍捡起的卡片选择位置，就只有一种方法来排列所有$\boldsymbol{13}$张卡片。对于每个$k$的值，有 $\binom{13}{k}-1$ 种方法选择第一遍的$k$个位置：我们排除那种第一遍就捡起所有13张卡片的排列。因此，答案是 \[\sum_{k=1}^{12}\left[\binom{13}{k}-1\right] = \left[\sum_{k=1}^{12}\binom{13}{k}\right]-12 = \left[\sum_{k=0}^{13}\binom{13}{k}\right]-14 = 2^{13} - 14 = \boxed{\textbf{(D) } 8178}\].

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