Difference between revisions of "2023 AIME I Problems/Problem 1"

Line 4: Line 4:
 
==Solution 1==
 
==Solution 1==
  
For simplicity purposes, we consider rotations as different arrangements.
+
For simplicity purposes, two arrangements are considered different even if they differ only by a rotation. Therefore, fourteen people have <math>14!</math> arrangements.
 +
 
 +
First, there are <math>\binom75</math> ways to choose <math>5</math> 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.
 +
 
 +
Together, the requested probability is <math></math>\frac{\binom75\cdot(10\cdot8\cdot6\cdot4\cdot2)\cdot9!}{14!} = \frac{48}{143},<math> from which the answer is </math>48+143 = \boxed{191}.<math>
  
We have <math>\binom 75</math> ways to choose <math>5</math> from the <math>7</math> diameters.
 
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
Line 13: Line 16:
 
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.
 
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 <math>1</math>st man anywhere on the table, now we have to place the <math>2</math>nd man somewhere around the table such that he is not diametrically opposite to the first man. This can happen with a probability of <math>\frac{12}{13}</math> because there are <math>13</math> available seats, and <math>12</math> of them are not opposite to the first man.
+
We first place the </math>1<math>st man anywhere on the table, now we have to place the </math>2<math>nd man somewhere around the table such that he is not diametrically opposite to the first man. This can happen with a probability of </math>\frac{12}{13}<math> because there are </math>13<math> available seats, and </math>12<math> of them are not opposite to the first man.
  
We do the same thing for the <math>3</math>rd man, finding a spot for him such that he is not opposite to the other <math>2</math> men, which would happen with a probability of <math>\frac{10}{12}</math> using similar logic. Doing this for the <math>4</math>th and <math>5</math>th men, we get probabilities of <math>\frac{8}{11}</math> and <math>\frac{6}{10}</math> respectively.
+
We do the same thing for the </math>3<math>rd man, finding a spot for him such that he is not opposite to the other </math>2<math> men, which would happen with a probability of </math>\frac{10}{12}<math> using similar logic. Doing this for the </math>4<math>th and </math>5<math>th men, we get probabilities of </math>\frac{8}{11}<math> and </math>\frac{6}{10}$ respectively.
  
 
Multiplying these probabilities, we get, <cmath>\frac{12}{13}\cdot\frac{10}{12}\cdot\frac{8}{11}\cdot\frac{6}{10}=\frac{48}{143}\longrightarrow\boxed{191}.</cmath>
 
Multiplying these probabilities, we get, <cmath>\frac{12}{13}\cdot\frac{10}{12}\cdot\frac{8}{11}\cdot\frac{6}{10}=\frac{48}{143}\longrightarrow\boxed{191}.</cmath>

Revision as of 14:17, 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 $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution 1

For simplicity purposes, two arrangements are considered different even if they differ only by a rotation. Therefore, fourteen people have $14!$ arrangements.

First, there are $\binom75$ ways to choose $5$ man-woman diameters. Then, there are $10\cdot8\cdot6\cdot4\cdot2$ ways to place the five men each in a man-woman diameter. Finally, there are $9!$ ways to place the nine women without restrictions.

Together, the requested probability is $$ (Error compiling LaTeX. Unknown error_msg)\frac{\binom75\cdot(10\cdot8\cdot6\cdot4\cdot2)\cdot9!}{14!} = \frac{48}{143},$from which the answer is$48+143 = \boxed{191}.$~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$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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