Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AIME II Problems/Problem 13"

(→‎Solution)
m (→‎Solution)
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
−
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{028}</math>.
+
Let the left shoes be <math>L1,\dots, L10</math> and the right shoes be <math>R1,\dots, R10.</math> Notice that there are <math>10 \cdot 9\cdot 8 \cdots 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\to R2</math>, <math>L2\to R5</math>, <math>L5 \to R1</math> is a cycle of length 3). Trivially, if there is a cycle of all ten shoes (for example, <math>L1 \to R2</math>, <math>L2 \to R3, \dots, L10 \to R1</math>) then there are <math>9 \cdot 8\cdots 1 = 9!</math> working cases. Or, there will be two cycles of 5 shoes: <math>\frac{{10\choose 5}}{2}\cdot{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 \cdot 24 \cdot 24}{10!} = \frac{1}{50}</math>. Adding the probability gives <math>\frac{3}{25}</math>, for an answer of <math>\boxed{028}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 19:35, 13 March 2015

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 $k<5$, no collection of $k$ pairs made by the child contains the shoes from exactly $k$ of the adults is $\frac{m}{n}$, where m and n are relatively prime positive integers. Find $m+n.$

Solution

Let the left shoes be $L1,\dots, L10$ and the right shoes be $R1,\dots, R10.$ Notice that there are $10 \cdot 9\cdot 8 \cdots 1 = 10!$ 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, $L1\to R2$, $L2\to R5$, $L5 \to R1$ is a cycle of length 3). Trivially, if there is a cycle of all ten shoes (for example, $L1 \to R2$, $L2 \to R3, \dots, L10 \to R1$) then there are $9 \cdot 8\cdots 1 = 9!$ working cases. Or, there will be two cycles of 5 shoes: $\frac{{10\choose 5}}{2}\cdot{4!}^2$. 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 $\frac{1}{10}$ and the second case $\frac{126 \cdot 24 \cdot 24}{10!} = \frac{1}{50}$. Adding the probability gives $\frac{3}{25}$, for an answer of $\boxed{028}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_13&oldid=68866"