Difference between revisions of "2014 AIME II Problems/Problem 13"
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==Solution== | ==Solution== | ||
| − | Let the left shoes be L1 ... L10 and the right shoes be R1 ... R10. Notice that there are 10 * 9 * 8 * ... * 1 = 10! possible pairings. To calculate the possible cases, we casework on the number and length of cycles, where a ''cycle'' is a sequence of shoes that starts with an arbitrary shoe, is continued when that shoe is paired with another shoe and the new shoe under consideration is the paired shoe's pair, and ends with the first shoe's pair. (For example, L1 -> R2, L2 -> R5, L5 -> R1 is a cycle of length 3). Trivially, if there is a cycle of all ten shoes (for example, L1 -> R2, L2 -> R3, ..., L10 -> R1) then there are 9 * 8 * ... * 1 = 9! working cases. Or, there will be two cycles of 5 shoes: <math>\frac{{10\choose 5}}{2}</math> * <math>{4!}^2</math>. Now, there cannot be a cycle of 6 and a cycle of 4 by the hypothesis; by similar reasoning, our two cases are the only ones that work. To compute the probability, we see that the first case yields a probability of 1/ | + | Let the left shoes be <math>L1 ... L10</math> and the right shoes be <math>R1 ... R10.</math> Notice that there are <math>10 * 9 * 8 * ... * 1 = 10!</math> possible pairings. To calculate the possible cases, we casework on the number and length of cycles, where a ''cycle'' is a sequence of shoes that starts with an arbitrary shoe, is continued when that shoe is paired with another shoe and the new shoe under consideration is the paired shoe's pair, and ends with the first shoe's pair. (For example, <math>L1</math> -> <math>R2</math>, <math>L2</math> -> <math>R5</math>, <math>L5</math> -> <math>R1</math> is a cycle of length 3). Trivially, if there is a cycle of all ten shoes (for example, <math>L1</math> -> <math>R2</math>, <math>L2</math> -> <math>R3, ..., L10</math> -> <math>R1</math>) then there are <math>9 * 8 * ... * 1 = 9!</math> working cases. Or, there will be two cycles of 5 shoes: <math>\frac{{10\choose 5}}{2}</math> * <math>{4!}^2</math>. Now, there cannot be a cycle of 6 and a cycle of 4 by the hypothesis; by similar reasoning, our two cases are the only ones that work. To compute the probability, we see that the first case yields a probability of <math>\frac{1}{10}</math> and the second case <math>\frac{126 * 24 * 24}{10!} = \frac{1}{50}</math>. Adding the probability gives <math>\frac{3}{25}</math>, for an answer of <math>\boxed{28}</math>. |
== See also == | == See also == | ||
Revision as of 00:16, 21 May 2014
Problem
Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer 
, no collection of 
pairs made by the child contains the shoes from exactly of the adults is 
, where m and n are relatively prime positive integers. Find 
Solution
Let the left shoes be 
and the right shoes be 
Notice that there are 
possible pairings. To calculate the possible cases, we casework on the number and length of cycles, where a cycle is a sequence of shoes that starts with an arbitrary shoe, is continued when that shoe is paired with another shoe and the new shoe under consideration is the paired shoe's pair, and ends with the first shoe's pair. (For example, 
-> 
, 
-> 
, 
-> 
is a cycle of length 3). Trivially, if there is a cycle of all ten shoes (for example, -> , -> 
-> ) then there are 
working cases. Or, there will be two cycles of 5 shoes: 
* 
. Now, there cannot be a cycle of 6 and a cycle of 4 by the hypothesis; by similar reasoning, our two cases are the only ones that work. To compute the probability, we see that the first case yields a probability of 
and the second case 
. Adding the probability gives 
, for an answer of 
.
See also
| 2014 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
