Difference between revisions of "2022 AMC 10A Problems/Problem 22"

Revision as of 18:40, 13 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?

$[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy]$ $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$

Solution

Solution 1

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

Solution 2

To solve this problem, we can use recursion on $n$ . Let $A_n$ be the number of arrangements for $n$ numbers. Now, let's look at how these arrangements are formed by case work on the first number $a_1$ .

If $a_1 = 1$ , the remaining $n-1$ numbers from $2$ to $n$ are arranged in the same way just like number 1 to $n-1$ in the case of $n-1$ numbers. So there are $A_{n-1}$ arrangements.

If $a_1 = 2$ , then we need to choose 1 position from position 2 to $n-1$ to put 1, and all remaining numbers must be arranged in increasing order, so there are $\binom{n-1}{1}$ such arrangements.

If $a_1 = k$ , then we need to choose $k-1$ positions from position 2 to $n-1$ to put $1, 2,\cdots k-1$ , and all remaining numbers must be arranged in increasing order, so there are $\binom{n-1}{k-1}$ such arrangements.

So we can write $A_n = A_{n-1} + \binom{n-1}{1} + \binom{n-1}{2} + \cdots + \binom{n-1}{n-1}$ which can be simplified to $A_n = A_{n-1} + 2^{n-1} - 1$ We can solve this recursive sequence by summing up $n-1$ lines of the recursive formula $A_n - A_{n-1} = 2^{n-1} - 1$ $A_{n-1} - A_{n-2} = 2^{n-2} - 1$ $\cdots$ $A_2 - A_{1} = 2^{1} - 1$ to get $A_n - A_1 = \sum_{k=1}^{n-1} (2^k - 1) = 2^n - 2 - (n-1) = 2^n - n - 1$ since $A_1=0$ , we have $A_n = 2^n - n - 1$ and $A_{13} = 2^{13} - 14 = \boxed{8178}$ .

So the answer is $\boxed{D}$

-- Dan Li

Solution 3

Let the cards picked up in the first pass be $1, 2, \ldots, n$ , where $1 \leq n \leq 12$ . Then there are $\binom{13}{n} -1$ ways the cards could have originally been arranged, as to place the cards we must choose $n$ of the original $13$ spots to place the $n$ cards from $1$ to $n$ (in that order). However, we must discard the case where the first $n$ spaces are taken up by these cards, as then the cards would all be picked up in one pass. Hence, our desired number of ways to arrange the cards is $\sum_{n=1}^{12} \left(\binom{13}{n} -1\right) = \left(\sum_{n=1}^{12} \binom{13}{n}\right) - 12 = \left(\sum_{n=0}^{13} \binom{13}{n}\right) - \binom{13}{0} - \binom{13}{13} - 12 = 2^{13} - 1 - 1 - 12 = \boxed{\textbf{(D)}\ 8178}.$

Solution 4 (Engineer's Induction, Risky)

When we have $3$ cards arranged in a row, after listing out all possible arrangements, we see that we have $4$ ones: $(1, 3, 2), (2, 1, 3,) (2, 3, 1),$ and $(3, 1, 2)$ . When we have $4$ cards, we find $11$ possible arrangements: $(1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (2, 1, 3, 4), (2, 3, 1, 4), (2, 3, 4, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 4, 1, 2),$ and $(4, 1, 2, 3).$ Hence, we recognize the pattern that for $n$ cards, we have $2^n - n - 1$ valid arrangements, so our answer is $2^{13} - 13 - 1 = \boxed{\textbf{(D)}\ 8178}.$ ~eibc

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 117: / Line 117: @@
 ~eibc
-== Video Solution By ThePuzzlr ==
+== Video Solution by ThePuzzlr ==
 https://youtu.be/p9xNduqTKLM

2022 AMC 10A (Problems • Answer Key • Resources)
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Difference between revisions of "2022 AMC 10A Problems/Problem 22"

Revision as of 18:40, 13 November 2022

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4 (Engineer's Induction, Risky)

Video Solution by ThePuzzlr

Solution by OmegaLearn Using Combinatorial Identities and Overcounting

Solution

See Also