2023 AIME I Problems/Problem 1
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[hide]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 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,
~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 5 men in a circle and the number of ways to arrange \binom{14}{5} = 2002.\binom{7}{2} = 212^{5} = 3221*32 = 672$valid arrangements.
Therefore, the probability that an arrangement is valid is$ (Error compiling LaTeX. Unknown error_msg)\frac{672}{2002} = \frac{48}{143}\boxed{191}.$
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 | ||
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