2023 AIME I Problems/Problem 1
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 
where 
and 
are relatively prime positive integers. Find 
Solution 1
For simplicity purposes, two arrangements are considered different even if they differ only by a rotation. Therefore, fourteen people have 
arrangements.
First, there are 
ways to choose 
man-woman diameters. Then, there are 
ways to place the five men each in a man-woman diameter. Finally, there are 
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},
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
2
\frac{12}{13}
13
12$of them are not opposite to the first man.
We do the same thing for the$ (Error compiling LaTeX. Unknown error_msg)3
2
\frac{10}{12}
4
5
\frac{8}{11}
\frac{6}{10}$ respectively.
Multiplying these probabilities, we get, ![\[\frac{12}{13}\cdot\frac{10}{12}\cdot\frac{8}{11}\cdot\frac{6}{10}=\frac{48}{143}\longrightarrow\boxed{191}.\]](http://latex.artofproblemsolving.com/e/3/6/e36fe62cd121ac6507c5b1d5bf944df760d10277.png)
~s214425
See also
| 2023 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
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