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| − | ==Problem==
| + | #redirect [[2022 AMC 10A Problems/Problem 22]] |
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| − | 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?
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| − | XXX [Image]
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| − | <math>\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191</math>
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| − | [[2022 AMC 10A Problems/Problem 22|Solution]] | |
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| − | ==Solution==
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| − | Since the <math>13</math> cards are picked up in two passes, the first pass must pick up the first <math>n</math> cards and the second pass must pick up the remaining cards <math>m</math> through <math>13</math>.
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| − | Also note that if <math>m</math>, which is the card that is numbered one more than <math>n</math>, is placed before <math>n</math>, then <math>m</math> will not be picked up on the first pass since cards are picked up in order. Therefore we desire <math>m</math> to be placed before <math>n</math> to create a second pass, and that after the first pass, the numbers <math>m</math> through <math>13</math> are lined up in order from least to greatest.
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| − | To construct this, <math>n</math> cannot go in the <math>n</math>th position because all cards <math>1</math> to <math>n-1</math> will have to precede it and there will be no room for <math>m</math>. Therefore <math>n</math> must be in slots <math>n+1</math> to <math>13</math>.
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| − | Let's do casework on which slot <math>n</math> goes into to get a general idea for how the problem works.
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| − | <math>\textbf{Case 1:}</math> With <math>n</math> in spot <math>n+1</math>, there are <math>n</math> available slots before <math>n</math>, and there are <math>n-1</math> cards preceding <math>n</math>. Therefore the number of ways to reserve these slots for the <math>n-1</math> cards is <math>\binom{n}{n-1}</math>. Then there is only <math>1</math> way to order these cards (since we want them in increasing order). Then card <math>m</math> goes into whatever slot is remaining, and the <math>13-m</math> cards are ordered in increasing order after slot <math>n+1</math>, giving only <math>1</math> way. Therefore in this case there are <math>\binom{n}{n-1}</math> possibilities.
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| − | <math>\textbf{Case 2:}</math> With <math>n</math> in spot <math>n+2</math>, there are <math>n+1</math> available slots before <math>n</math>, and there are <math>n-1</math> cards preceding <math>n</math>. Therefore the number of ways to reserve slots for these cards are <math>\binom{n+1}{n-1}</math>. Then there is one way to order these cards. Then cards <math>m</math> and <math>m+1</math> 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 <math>m+2</math> to <math>13</math> will be ordered in increasing order after slot <math>n+1</math>, which yields <math>1</math> way. Therefore, this case has <math>\binom{n+1}{n-1}</math> possibilities.
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| − | I think we can see a general pattern now. With <math>n</math> in slot <math>x</math>, there are <math>x-1</math> slots to distribute to the previous <math>n-1</math> cards, which can be done in <math>\binom{x-1}{n-1}</math> ways. Then the remaining cards fill in in just <math>1</math> way. Since the cases of <math>n</math> start in slot <math>n+1</math> and end in slot <math>13</math>, this sum amounts to:
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| − | <cmath>\binom{n}{n-1}+\binom{n+1}{n-1}+\binom{n+2}{n-1} + \cdots + \binom{12}{n-1}</cmath> for any <math>n</math>.
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| − | Hmmm... where have we seen this before?
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| − | We use wishful thinking to add a term of <math>\binom{n-1}{n-1}</math>:
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| − | <cmath>\binom{n-1}{n-1}+\binom{n}{n-1}+\binom{n+1}{n-1}+\binom{n+2}{n-1} + \cdots + \binom{12}{n-1}</cmath>
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| − | This is just the hockey stick identity! Applying it, this expression is equal to <math>\binom{13}{n}</math>. However, we added an extra term, so subtracting it off, the total number of ways to order the <math>13</math> cards for any <math>n</math> is <cmath>\binom{13}{n}-1</cmath>
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| − | Finally, to calculate the total for all <math>n</math>, we sum from <math>n=0</math> to <math>13</math>. This yields us:
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| − | <cmath>\sum_{n=0}^{13} \binom{13}{n}-1 \implies \sum_{n=0}^{13} \binom{13}{n} - \sum_{n=0}^{13} 1</cmath>
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| − | <cmath>\implies 2^{13} - 14 = 8192 - 14 = 8178 = \boxed{D}</cmath>
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| − | ~KingRavi
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| − | == Video Solution By ThePuzzlr ==
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| − | https://youtu.be/p9xNduqTKLM
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| − | ~ MathIsChess
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| − | ==Solution by OmegaLearn Using Combinatorial Identities and Overcounting==
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| − | https://youtu.be/gW8gPEEHSfU
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| − | ~ pi_is_3.14
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| − | ==Solution==
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| − | https://youtu.be/ZGqrs5eg6-s
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| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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| − | == See Also ==
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| − | {{AMC12 box|year=2022|ab=A|num-b=18|num-a=20}}
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| − | {{AMC10 box|year=2022|ab=A|num-b=21|num-a=23}}
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| − | {{MAA Notice}}
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