Difference between revisions of "2023 AIME I Problems/Problem 1"
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We can reword the problem into a simpler problem to solve. The new problem is "What is the probability two men don't stand opposite of each other?" We will order the men first, because the identities of each man and women don't matter. The first man can stand anywhere (<math>\frac{14}{14}</math>), then the second man can stand in 12 spaces out of 13, third man can stand in 10 places, so on until the fifth man stands in 6 places out of 10 possible places. <cmath>\frac{14}{14}\cdot\frac{12}{13}\cdot\frac{10}{12}+\frac{8}{11}+\frac{6}{10} = \frac{48}{143}</cmath> | We can reword the problem into a simpler problem to solve. The new problem is "What is the probability two men don't stand opposite of each other?" We will order the men first, because the identities of each man and women don't matter. The first man can stand anywhere (<math>\frac{14}{14}</math>), then the second man can stand in 12 spaces out of 13, third man can stand in 10 places, so on until the fifth man stands in 6 places out of 10 possible places. <cmath>\frac{14}{14}\cdot\frac{12}{13}\cdot\frac{10}{12}+\frac{8}{11}+\frac{6}{10} = \frac{48}{143}</cmath> | ||
This gives <math>m+n = \boxed{191}.</math> | This gives <math>m+n = \boxed{191}.</math> | ||
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~chem1kall | ~chem1kall | ||
Revision as of 08:42, 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 
Solutions
Solution 1
We can reword the problem into a simpler problem to solve. The new problem is "What is the probability two men don't stand opposite of each other?" We will order the men first, because the identities of each man and women don't matter. The first man can stand anywhere (
), then the second man can stand in 12 spaces out of 13, third man can stand in 10 places, so on until the fifth man stands in 6 places out of 10 possible places. ![\[\frac{14}{14}\cdot\frac{12}{13}\cdot\frac{10}{12}+\frac{8}{11}+\frac{6}{10} = \frac{48}{143}\]](http://latex.artofproblemsolving.com/1/9/8/198bd49815708bbbc4ed1dc9acc2ca402c16ac68.png)
This gives 
~chem1kall
Solution 2
Something else
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 | ||
