Difference between revisions of "2023 AIME I Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
| − | For simplicity purposes, arrangements | + | 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. |
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 18:01, 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 
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. 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
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
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. 
