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Difference between revisions of "2023 AIME I Problems/Problem 1"

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==Solution 3==
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Assume that rotations and reflections are distinct arrangements, and replace men and women with identical M's and W's, respectively. (We can do that because the number of ways to arrange 5 men in a circle and the number of ways to arrange <math>94 women in a circle, are constants.) The total number of ways to arrange 5 M's and 9 W's is </math>\binom{14}{5} = 2002.<math> 
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To count the number of valid arrangements, we notice that exactly 2 of the pairs of diametrically opposite positions must be occupied by 2 women. There are </math>\binom{7}{2} = 21<math> ways to choose these 2 pairs. For the remaining 5 pairs, we have to choose which position is occupied by a man and which is occupied by a woman. This can be done in </math>2^{5} = 32<math> ways. Therefore, there are </math>21*32 = 672<math> valid arrangements.
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Therefore, the probability that an arrangement is valid is </math>\frac{672}{2002} = \frac{48}{143}<math> for an answer of </math>\boxed{191}.$
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pianoboy
  
 
==See also==
 
==See also==
 
{{AIME box|year=2023|before=First Problem|num-a=2|n=I}}
 
{{AIME box|year=2023|before=First Problem|num-a=2|n=I}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:27, 8 February 2023

Problem

Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution 1

For simplicity purposes, arrangements that differ only by a rotation are considered different. So, there are $14!$ arrangements without restrictions.

First, there are $\binom75$ ways to choose the man-woman diameters. Then, there are $10\cdot8\cdot6\cdot4\cdot2$ ways to place the five men each in a man-woman diameter. Finally, there are $9!$ ways to place the nine women without restrictions.

Together, the requested probability is \[\frac{\tbinom75\cdot(10\cdot8\cdot6\cdot4\cdot2)\cdot9!}{14!} = \frac{48}{143},\] from which the answer is $48+143 = \boxed{191}.$

~MRENTHUSIASM

Solution 2

This problem is equivalent to solving for the probability that no man is standing diametrically opposite to another man. We can simply just construct this.

We first place the $1$st man anywhere on the circle, now we have to place the $2$nd man somewhere around the circle such that he is not diametrically opposite to the first man. This can happen with a probability of $\frac{12}{13}$ because there are $13$ available spots, and $12$ of them are not opposite to the first man.

We do the same thing for the $3$rd man, finding a spot for him such that he is not opposite to the other $2$ men, which would happen with a probability of $\frac{10}{12}$ using similar logic. Doing this for the $4$th and $5$th men, we get probabilities of $\frac{8}{11}$ and $\frac{6}{10}$ respectively.

Multiplying these probabilities, we get, \[\frac{12}{13}\cdot\frac{10}{12}\cdot\frac{8}{11}\cdot\frac{6}{10}=\frac{48}{143}\longrightarrow\boxed{191}.\]

~s214425

Solution 3

Assume that rotations and reflections are distinct arrangements, and replace men and women with identical M's and W's, respectively. (We can do that because the number of ways to arrange 5 men in a circle and the number of ways to arrange $94 women in a circle, are constants.) The total number of ways to arrange 5 M's and 9 W's is$\binom{14}{5} = 2002.$To count the number of valid arrangements, we notice that exactly 2 of the pairs of diametrically opposite positions must be occupied by 2 women. There are$\binom{7}{2} = 21$ways to choose these 2 pairs. For the remaining 5 pairs, we have to choose which position is occupied by a man and which is occupied by a woman. This can be done in$2^{5} = 32$ways. Therefore, there are$21*32 = 672$valid arrangements.

Therefore, the probability that an arrangement is valid is$ (Error compiling LaTeX. Unknown error_msg)\frac{672}{2002} = \frac{48}{143}$for an answer of$\boxed{191}.$

pianoboy

See also

2023 AIME I (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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