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Revision as of 15:48, 8 February 2023
Contents
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, arrangements that differ only by a rotation are considered different. So, there are 
arrangements without restrictions.
First, there are 
ways to choose the 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 ![\[\frac{\tbinom75\cdot(10\cdot8\cdot6\cdot4\cdot2)\cdot9!}{14!} = \frac{48}{143},\]](http://latex.artofproblemsolving.com/3/a/a/3aaddce0b4fceb9983f29b56296910999b3a0e53.png)
from which the answer is 
~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 
st man anywhere on the circle, now we have to place the 
nd man somewhere around the circle such that he is not diametrically opposite to the first man. This can happen with a probability of 
because there are 
available spots, and 
of them are not opposite to the first man.
We do the same thing for the 
rd man, finding a spot for him such that he is not opposite to the other men, which would happen with a probability of 
using similar logic. Doing this for the 
th and 
th men, we get probabilities of 
and 
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 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
