Difference between revisions of "2023 AIME I Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
| − | For simplicity purposes, we | + | For simplicity purposes, we consider rotations as different arrangements. |
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| + | We have <math>\binom 75</math> ways to choose <math>5</math> from the <math>7</math> diameters. | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 13:08, 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, we consider rotations as different arrangements.
We have 
ways to choose 
from the 
diameters.
~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 
st man anywhere on the table, now we have to place the 
nd man somewhere around the table such that he is not diametrically opposite to the first man. This can happen with a probability of 
because there are 
available seats, 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 Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
