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

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==Problem==
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#redirect [[2022 AMC 10A Problems/Problem 22]]
 
 
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 <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>.
 
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.
 
 
 
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>.
 
Let's do casework on which slot <math>n</math> goes into to get a general idea for how the problem works.
 
With <math>n</math> in spot <math>n+1</math>,
 
 
 
 
 
Solution in Progress
 
 
 
~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)
 
 
 
== See Also ==
 
 
 
{{AMC12 box|year=2022|ab=A|num-b=18|num-a=20}}
 
{{AMC10 box|year=2022|ab=A|num-b=21|num-a=23}}
 
{{MAA Notice}}
 

Latest revision as of 19:38, 13 November 2022