Difference between revisions of "2023 AIME I Problems/Problem 1"
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− | Assume that rotations and reflections are distinct arrangements, and replace men and women with identical M's and W's, respectively. (We can do that because the number of ways to arrange <math>5</math> men in a circle and the number of ways to arrange <math> | + | Assume that rotations and reflections are distinct arrangements, and replace men and women with identical M's and W's, respectively. (We can do that because the number of ways to arrange <math>5</math> men in a circle and the number of ways to arrange <math>9</math> women in a circle, are constants.) The total number of ways to arrange <math>5</math> M's and <math>9</math> W's is <math>\binom{14}{5} = 2002.</math> |
To count the number of valid arrangements (i.e. arrangements where every M is diametrically opposite a W), we notice that exactly <math>2</math> of the pairs of diametrically opposite positions must be occupied by <math>2</math> W's. There are <math>\binom{7}{2} = 21</math> ways to choose these <math>2</math> pairs. For the remaining <math>5</math> pairs, we have to choose which position is occupied by an M and which is occupied by a W. This can be done in <math>2^{5} = 32</math> ways. Therefore, there are <math>21*32 = 672</math> valid arrangements. | To count the number of valid arrangements (i.e. arrangements where every M is diametrically opposite a W), we notice that exactly <math>2</math> of the pairs of diametrically opposite positions must be occupied by <math>2</math> W's. There are <math>\binom{7}{2} = 21</math> ways to choose these <math>2</math> pairs. For the remaining <math>5</math> pairs, we have to choose which position is occupied by an M and which is occupied by a W. This can be done in <math>2^{5} = 32</math> ways. Therefore, there are <math>21*32 = 672</math> valid arrangements. |
Revision as of 10:55, 10 February 2023
Contents
[hide]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, we consider two arrangements different even if they only differ by rotations or reflections. In this way, there are arrangements without restrictions.
First, there are ways to choose the 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 from which the answer is
~MRENTHUSIASM
Solution 2
We can simply just loop through each of the men and find the probability that the person opposite from him is a woman.
Start by sitting down the st man. The probability that the person opposite to him is a woman is since out of the people who can sit opposite to him, can be a woman. With the nd man, we can use the same logic: there are people who can sit opposite to him, but only of them are a woman, so the probability is We use the same logic for the rd, th and th men to get probabilities of , and respectively.
Multiplying these probabilities, we get a final answer of
~s214425
Solution 3
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 st man anywhere on the circle, now we have to place the nd man somewhere around the circle such that he is not diametrically opposite to the first man. This can happen with a probability of because there are available spots, 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,
~s214425
Solution 4
Assume that rotations and reflections are distinct arrangements, and replace men and women with identical M's and W's, respectively. (We can do that because the number of ways to arrange men in a circle and the number of ways to arrange women in a circle, are constants.) The total number of ways to arrange M's and W's is
To count the number of valid arrangements (i.e. arrangements where every M is diametrically opposite a W), we notice that exactly of the pairs of diametrically opposite positions must be occupied by W's. There are ways to choose these pairs. For the remaining pairs, we have to choose which position is occupied by an M and which is occupied by a W. This can be done in ways. Therefore, there are valid arrangements.
Therefore, the probability that an arrangement is valid is for an answer of
~pianoboy
Solution 5
To start off, we calculate the total amount of ways to organize all people irrespective of any constraints. This is simply , because we just count how many ways we can place all men in any of the slots.
Since men cannot be diametrically opposite with each other, because of the constraints, placing down one man in any given spot will make another spot on the opposite side of the circle unable to hold any men. This means that placing down one man will effectively take away spots.
There are possible slots the first man can be placed. Once that man was placed, the next man only have possible slots because the slot that the first man is in is taken and the diametrically opposite spot to the first man can't have any men. Similar logic applies for the third man, who has possible slots. The fourth man has possible slots, and the fifth man has possible slots.
This means the number of ways you can place all men down is . However, since the men are all indistinct from each other, you also have to divide that value by , since there are ways to arrange the men in each possible positioning of the men on the circle. This means the total number of ways to arrange the men around the circle so that none of them are diametrically opposite of each other is: . The women simply fill in the rest of the available slots in each arrangement of men.
Thus, the final probability is , meaning the answer is .
~ericshi1685
Solution 6
We will first assign seats to the men. The first man can be placed in any of the slots. The second man can be placed in any of the remaining seats, except for the one diametrically opposite to the first man. So, there are ways to seat him. With a similar argument, the third man can be seated in ways, the fourth man in ways and the last man in ways.
So, the total number of ways to arrange the men is .
The women go to the remaining 9 spots. Note that since none of the seats diametrically opposite to the men is occupied, each man is opposite a woman. The number of ways to arrange the women is therefore, simply , meaning that the total number of ways to arrange the people with restrictions is In general, there are ways to arrange the people without restrictions. So, the probability is The answer is .
~baassid24
Solution 7
First pin one man on one seat (to ensure no rotate situations). Then there are arrangements. Because 5 men must have women at their opposite side, we consider the 2nd man and the woman opposite as one group and name it . There are 4 groups, except the first man pinned on the same point. And for the rest 4 women, name them and . First to order , there are ways. For the 1st man, there are 9 women to choose, 8 for the 2nd, , 5 for the 5th, and then for the 2 women pairs 3 and 1. Because every 2 person in the group have chance to change their position, there are possibilities.
So the possibility is
The answer is
~PLASTA
Solution 8
We get around the condition that each man can't be opposite to another man by simply considering all diagonals, and choosing where there will be a single man. For each diagonal, the man can go on either side, and there are ways to arrange the men and the women in total. Thus our answer is We get
~AtharvNaphade
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=KdKysmdgepI
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
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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