Difference between revisions of "2023 AIME I Problems/Problem 1"
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Use combinatorics | Use combinatorics | ||
| − | ===Solution 2=== | + | ===Solution 2 (constructive)=== |
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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. | ||
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| + | 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. | ||
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| + | 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. | ||
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| + | 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> | ||
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| + | ~s214425 | ||
==See also== | ==See also== | ||
{{AIME box|year=2023|before=First Question|num-a=2|n=I}} | {{AIME box|year=2023|before=First Question|num-a=2|n=I}} | ||
Revision as of 12:29, 8 February 2023
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 
be the probability that every man stands diametrically opposite from a woman.
Find 
.
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 
st man anywhere on the table, now we have to place the 
nd man somewhere around the table such that he is not diametrically opposite to the first man. This can happen with a probability of 
because there are 
available seats, and 
of them are not opposite to the first man.
We do the same thing for the 
rd man, finding a spot for him such that he is not opposite to the other men, which would happen with a probability of 
using similar logic. Doing this for the 
th and 
th men, we get probabilities of 
and 
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 | ||
