2023 AIME I Problems/Problem 1

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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, two arrangements are considered different even if they differ only by a rotation. Therefore, fourteen people have $14!$ arrangements.

First, there are $\binom75$ ways to choose $5$ 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 $$ (Error compiling LaTeX. Unknown error_msg)\frac{\binom75\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 sitting diametrically opposite to another man. We can simply just construct this.

We first place the$ (Error compiling LaTeX. Unknown error_msg)1$st man anywhere on the table, now we have to place the$2$nd man somewhere around the table 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 seats, and$12$of them are not opposite to the first man.

We do the same thing for the$ (Error compiling LaTeX. Unknown error_msg)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

See also

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