Difference between revisions of "2023 AIME I Problems/Problem 1"
MRENTHUSIASM (talk | contribs) |
MRENTHUSIASM (talk | contribs) |
||
| Line 4: | Line 4: | ||
==Solution 1== | ==Solution 1== | ||
| − | For simplicity purposes, | + | 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> | ||
| − | |||
~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< | + | 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< | + | 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 13:17, 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 
where 
and 
are relatively prime positive integers. Find 
Solution 1
For simplicity purposes, two arrangements are considered different even if they differ only by a rotation. Therefore, fourteen people have 
arrangements.
First, there are 
ways to choose 
man-woman diameters. Then, there are 
ways to place the five men each in a man-woman diameter. Finally, there are 
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},
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
2
\frac{12}{13}
13
12$of them are not opposite to the first man.
We do the same thing for the$ (Error compiling LaTeX. Unknown error_msg)3
2
\frac{10}{12}
4
5
\frac{8}{11}
\frac{6}{10}$ respectively.
Multiplying these probabilities, we get, ![\[\frac{12}{13}\cdot\frac{10}{12}\cdot\frac{8}{11}\cdot\frac{6}{10}=\frac{48}{143}\longrightarrow\boxed{191}.\]](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 | ||
