Difference between revisions of "2022 AMC 12A Problems/Problem 19"

Revision as of 12:08, 12 November 2022

Problem

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?

XXX [Image]

$\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$

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)

@@ Line 1: / Line 1: @@
 ==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.
+Suppose that 13 cards numbered <math>1, 2, 3, \cdots, 13</math> 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 <math>13!</math> possible orderings of the cards will the <math>13</math> cards be picked up in exactly two passes?
-For how many of the 13! possible orderings of the cards will the 13 cards be picked up in exactly
+XXX [Image]
-two passes?
+<math>\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191</math>
+[[2022 AMC 10A Problems/Problem 22|Solution]]
 ==Solution==

2022 AMC 12A (Problems • Answer Key • Resources)
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2022 AMC 10A (Problems • Answer Key • Resources)
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Difference between revisions of "2022 AMC 12A Problems/Problem 19"

Revision as of 12:08, 12 November 2022

Contents

Problem

Solution

Video Solution By ThePuzzlr

Solution by OmegaLearn Using Combinatorial Identities and Overcounting

Solution

See Also