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, we consider two arrangements different even if they only differ by rotations or reflections. In this way, 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
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 men in a circle and the number of ways to arrange 
women in a circle, are constants.) The total number of ways to arrange M's and 
W's is 
To count the number of valid arrangements (i.e. arrangements where every M is diametrically opposite a W), we notice that exactly of the pairs of diametrically opposite positions must be occupied by W's. There are 
ways to choose these pairs. For the remaining pairs, we have to choose which position is occupied by an M and which is occupied by a W. This can be done in 
ways. Therefore, there are 
valid arrangements.
Therefore, the probability that an arrangement is valid is 
for an answer of 
pianoboy
Solution 4
First, we calculate the total amount of ways to organize all 
people irrespective of any constraints. This is simply 
, because we just count how many ways we can place all men in any of the "slots".
Since men cannot be diametrically opposite with each other, because of the constraints, placing down one man in any given spot will make another spot on the opposite side of the circle only able to hold women and no men. This means that placing down one man will effectively take away spots.
There are possible slots the first man can be placed. Once that man was placed, the next man only have possible slots because the slot that the first man is in is taken and the diametrically opposite spot to the first man can't have any men. Similar logic applies for the third man, who has 
possible slots. The fourth man has 
possible slots, and the fifth man has 
possible slots.
This means the number of ways you can place all men down is 
. However, since the men are all indistinct from each other, you also have to divide that value by 
, since there are 
ways to arrange the men in each possible positioning of the men on the circle. This means the total number of ways to arrange the men around the circle so that none of them are diametrically opposite of each other is: 
. The women simply fill in the rest of the available slots in each arrangement of men.
Thus, the final probability is , meaning the answer is 
.
~ericshi1685
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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