2022 AMC 12A Problems/Problem 19

Problem

Suppose that 13 cards numbered 1, 2, 3, . . . , 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?

Solution

Since the $13$ cards are picked up in two passes, the first pass must pick up the first $n$ cards and the second pass must pick up the remaining cards $m$ through $13$ . Also note that if $m$ , which is the card that is numbered one more than $n$ , is placed before $n$ , then $m$ will not be picked up on the first pass since cards are picked up in order. Therefore we desire $m$ to be placed before $n$ to create a second pass, and that after the first pass, the numbers $m$ through $13$ are lined up in order from least to greatest.

To construct this, $n$ cannot go in the $n$ th position because all cards $1$ to $n-1$ will have to precede it and there will be no room for $m$ . Therefore $n$ must be in slots $n+1$ to $13$ . Let's do casework on which slot $n$ goes into to get a general idea for how the problem works.

$\textbf{Case 1:}$ With $n$ in spot $n+1$ , there are $n$ available slots before $n$ , and there are $n-1$ cards preceding $n$ . Therefore the number of ways to reserve these slots for the $n-1$ cards is $\binom{n}{n-1}$ . Then there is only $1$ way to order these cards (since we want them in increasing order). Then card $m$ goes into whatever slot is remaining, and the $13-m$ cards are ordered in increasing order after slot $n+1$ , giving only $1$ way. Therefore in this case there are $\binom{n}{n-1}$ possibilities.

$\textbf{Case 2:}$ With $n$ in spot $n+2$ , there are $n+1$ available slots before $n$ , and there are $n-1$ cards preceding $n$ . Therefore the number of ways to reserve slots for these cards are $\binom{n+1}{n-1}$ . Then there is one way to order these cards. Then cards $m$ and $m+1$ must go in the remaining two slots, and there is only one way to order them since they must be in increasing order. Finally, cards $m+2$ to $13$ will be ordered in increasing order after slot $n+1$ , which yields $1$ way. Therefore, this case has $\binom{n+1}{n-1}$ possibilities.

I think we can see a general pattern now. With $n$ in slot $x$ , there are $x-1$ slots to distribute to the previous $n-1$ cards, which can be done in $\binom{x-1}{n-1}$ ways. Then the remaining cards fill in in just $1$ way. Since the cases of $n$ start in slot $n+1$ and end in slot $13$ , this sum amounts to: $\binom{n}{n-1}+\binom{n+1}{n-1}+\binom{n+2}{n-1} + \cdots + \binom{12}{n-1}$ for any $n$ .

Hmmm... where have we seen this before?

We use wishful thinking to add a term of $\binom{n-1}{n-1}$ : $\binom{n-1}{n-1}+\binom{n}{n-1}+\binom{n+1}{n-1}+\binom{n+2}{n-1} + \cdots + \binom{12}{n-1}$

This is just the hockey stick identity! Applying it, this expression is equal to $\binom{13}{n}$ . However, we added an extra term, so subtracting it off, the total number of ways to order the $13$ cards for any $n$ is $\binom{13}{n}-1$

Finally, to calculate the total for all $n$ , we sum from $n=0$ to $13$ . This yields us:

$\sum_{n=0}{13} \binom{13}{n}-1 \implies \sum_{n=0}{13} \binom{13}{n} - \sum_{n=0}{13} -1$ $\implies 2^{13} - 14 = 8192 - 14 = 8178 = \boxed{D}$

~KingRavi

Video Solution By ThePuzzlr

https://youtu.be/p9xNduqTKLM

~ MathIsChess

Solution by OmegaLearn Using Combinatorial Identities and Overcounting

https://youtu.be/gW8gPEEHSfU

~ pi_is_3.14

Solution

https://youtu.be/ZGqrs5eg6-s

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 12 Problems and Solutions

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2022 AMC 12A Problems/Problem 19

Contents

Problem

Solution

Video Solution By ThePuzzlr

Solution by OmegaLearn Using Combinatorial Identities and Overcounting

Solution

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