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Difference between revisions of "2023 AIME I Problems/Problem 1"
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==Solution 1==
 
==Solution 1==
  
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For simplicity purposes, two arrangements are considered different even if they differ only by a rotation. Therefore, fourteen people have <math>14!</math> arrangements.
+
For simplicity purposes, arrangements that differ only by a rotation are considered different. So, there are <math>14!</math> arrangements without restrictions.
  
−
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.
+
First, there are <math>\binom75</math> ways to choose the 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>
+
Together, the requested probability is <cmath>\frac{\tbinom75\cdot(10\cdot8\cdot6\cdot4\cdot2)\cdot9!}{14!} = \frac{48}{143},</cmath> from which the answer is <math>48+143 = \boxed{191}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
Line 14: Line 14:
 
==Solution 2==
 
==Solution 2==
  
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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 standing 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 circle, now we have to place the <math>2</math>nd man somewhere around the circle 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 spots, 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}$ 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}</math> 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 13:23, 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, arrangements that differ only by a rotation are considered different. So, there are $14!$ arrangements without restrictions.

First, there are $\binom75$ ways to choose the 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 \[\frac{\tbinom75\cdot(10\cdot8\cdot6\cdot4\cdot2)\cdot9!}{14!} = \frac{48}{143},\] 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 standing diametrically opposite to another man. We can simply just construct this.

We first place the $1$st man anywhere on the circle, now we have to place the $2$nd man somewhere around the circle 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 spots, 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 (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
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