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

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==Solution 1==
 
==Solution 1==
  
For simplicity purposes, we consider two arrangements different even if they only differ by rotations or reflections. So, there are <math>14!</math> arrangements without restrictions.
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For simplicity purposes, we consider two arrangements different even if they differ by rotations or reflections. So, there are <math>14!</math> arrangements without restrictions.
  
 
First, there are <math>\binom75</math> ways to choose the man-woman diameters. Then, there are <math>10\cdot8\cdot6\cdot4\cdot2</math> ways to place the five men each in a man-woman diameter. Finally, there are <math>9!</math> ways to place the nine women without restrictions.
 
First, there are <math>\binom75</math> ways to choose the man-woman diameters. Then, there are <math>10\cdot8\cdot6\cdot4\cdot2</math> ways to place the five men each in a man-woman diameter. Finally, there are <math>9!</math> ways to place the nine women without restrictions.

Revision as of 18:06, 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, we consider two arrangements different even if they differ by rotations or reflections. 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 (i.e. arrangements where every M is diametrically opposite a W), we notice that exactly $2$ of the pairs of diametrically opposite positions must be occupied by $2$ W's. 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 an M and which is occupied by a W. 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 $\frac{672}{2002} = \frac{48}{143}$ for an answer of $\boxed{191}.$

pianoboy

See also

2023 AIME I (ProblemsAnswer KeyResources)
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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