Question

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?

$4082$

$4095$

$4096$

$8178$

$8191$

Solution:

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 $\sum_{n=1}^{12}\left(\binom{13}{n}-1\right)=(2^{13}-2)-12=8178.$

Thus, D is the correct answer.

2022 AMC 10A Problem 22

Solution:

Problem 22 in Other Years