Difference between revisions of "2022 AMC 12A Problems/Problem 19"
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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. | Let's do casework on which slot <math>n</math> goes into to get a general idea for how the problem works. | ||
| − | \textbf{Case 1:}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). Finally, card <math>m</math> goes into whatever slot is remaining. Therefore in this case there are <math>\binom{n}{n-1}</math> possibilities. | + | <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). Finally, card <math>m</math> goes into whatever slot is remaining. Therefore in this case there are <math>\binom{n}{n-1}</math> possibilities. |
Revision as of 11:52, 12 November 2022
Contents
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 
cards are picked up in two passes, the first pass must pick up the first 
cards and the second pass must pick up the remaining cards 
through .
Also note that if , which is the card that is numbered one more than , is placed before , then will not be picked up on the first pass since cards are picked up in order. Therefore we desire to be placed before to create a second pass, and that after the first pass, the numbers through are lined up in order from least to greatest.
To construct this, cannot go in the th position because all cards 
to 
will have to precede it and there will be no room for . Therefore must be in slots 
to .
Let's do casework on which slot goes into to get a general idea for how the problem works.
With in spot , there are available slots before , and there are cards preceding . Therefore the number of ways to reserve these slots for the cards is 
. Then there is only way to order these cards (since we want them in increasing order). Finally, card goes into whatever slot is remaining. Therefore in this case there are possibilities.
Solution in Progress
~KingRavi
Video Solution By ThePuzzlr
~ MathIsChess
Solution by OmegaLearn Using Combinatorial Identities and Overcounting
~ pi_is_3.14
Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
| 2022 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
