Question

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?

$4082$

$4095$

$4096$

$8178$

$8191$

Answer: D

Concepts:permutations pattern recognition

Difficulty rating: 2010

Solution:

Let $\text{pos}(k)$ be the position of card $k.$ A fresh pass is needed exactly when $\text{pos}(k+1)\lt\text{pos}(k),$ so the number of passes is one more than the number of descents in the sequence $\text{pos}(1),\text{pos}(2),\ldots,\text{pos}(13).$

To build a permutation with at most one descent, choose the entries before the possible descent and write both chosen blocks in increasing order. There are $2^{13}$ subsets. The $14$ initial segments $\varnothing,\{1\},\ldots,\{1,\ldots,13\}$ produce no descent; every other subset produces a unique permutation with one descent. Hence the count is $2^{13}-14=8178.$

Thus, the correct answer is D.

2022 AMC 12A Problem 19

Answer: D

Difficulty rating: 2010

Solution:

Problem 19 in Other Years