2023 AIME I Problems/Problem 1

Revision as of 12:29, 8 February 2023 by S214425 (talk | contribs) (Solution 2)

Problem

Note: This is not official (I am posting this because the AIME I is officially over). Please post official problem statement after it is released.

There are five men and nine women randomly arranged in a circle. Let $\frac{x}{y}$ be the probability that every man stands diametrically opposite from a woman.

Find $x+y$.

Solutions

Solution 1

Use combinatorics

Solution 2 (constructive)

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 $1$st man anywhere on the table, now we have to place the $2$nd man somewhere around the table such that he is not diametrically opposite to the first man. This can happen with a probability of $\frac{12}{13}$ because there are $13$ available seats, and $12$ of them are not opposite to the first man.

We do the same thing for the $3$rd man, finding a spot for him such that he is not opposite to the other $2$ men, which would happen with a probability of $\frac{10}{12}$ using similar logic. Doing this for the $4$th and $5$th men, we get probabilities of $\frac{8}{11}$ and $\frac{6}{10}$ respectively.

Multiplying these probabilities, we get, \[\frac{12}{13}\cdot\frac{10}{12}\cdot\frac{8}{11}\cdot\frac{6}{10}=\frac{48}{143}\longrightarrow\boxed{191}.\]

~s214425

See also

2023 AIME I (ProblemsAnswer KeyResources)
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