Difference between revisions of "2023 AIME I Problems/Problem 1"
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==Problem== | ==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 <math>\frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math> | 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 <math>\frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math> | ||
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For simplicity purposes, we consider two arrangements different even if they only differ by rotations or reflections. In this way, there are <math>14!</math> arrangements without restrictions. | For simplicity purposes, we consider two arrangements different even if they only differ by rotations or reflections. In this way, there are <math>14!</math> arrangements without restrictions. | ||
− | First, there are <math>\ | + | First, there are <math>\binom{7}{5}</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 <cmath>\frac{\ | + | Together, the requested probability is <cmath>\frac{\tbinom{7}{5}\cdot(10\cdot8\cdot6\cdot4\cdot2)\cdot9!}{14!} = \frac{21\cdot(10\cdot8\cdot6\cdot4\cdot2)}{14\cdot13\cdot12\cdot11\cdot10} = \frac{48}{143},</cmath> from which the answer is <math>48+143 = \boxed{191}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
==Solution 2== | ==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 <math>1</math>st man. The probability that the person opposite to him is a woman is <math>\frac{9}{13}</math> since out of the <math>13</math> people who can sit opposite to him, <math>9</math> can be a woman. With the <math>2</math>nd man, we can use the same logic: there are <math>11</math> people who can sit opposite to him, but only <math>8</math> of them are a woman, so the probability is <math>\frac{8}{11}.</math> We use the same logic for the <math>3</math>rd, <math>4</math>th and <math>5</math>th men to get probabilities of <math>\frac{7}{9}</math>, <math>\frac{6}{7}</math> and <math>\frac{5}{5},</math> respectively. | ||
+ | |||
+ | Multiplying these probabilities, we get a final answer of <cmath>\frac{9}{13}\cdot\frac{8}{11}\cdot\frac{7}{9}\cdot\frac{6}{7}\cdot\frac{5}{5}=\frac{48}{143}\longrightarrow\boxed{191}.</cmath> | ||
+ | |||
+ | ~s214425 (Inspired by Math Jam) | ||
+ | |||
+ | ==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. | 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. | ||
Line 24: | Line 35: | ||
~s214425 | ~s214425 | ||
− | ==Solution | + | ==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 <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. | ||
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Therefore, the probability that an arrangement is valid is <math>\frac{672}{2002} = \frac{48}{143}</math> for an answer of <math>\boxed{191}.</math> | Therefore, the probability that an arrangement is valid is <math>\frac{672}{2002} = \frac{48}{143}</math> for an answer of <math>\boxed{191}.</math> | ||
− | pianoboy | + | ~pianoboy |
− | ==Solution | + | ==Solution 5== |
To start off, we calculate the total amount of ways to organize all <math>14</math> people irrespective of any constraints. This is simply <math>{14\choose5} = 2002</math>, because we just count how many ways we can place all <math>5</math> men in any of the <math>14</math> slots. | To start off, we calculate the total amount of ways to organize all <math>14</math> people irrespective of any constraints. This is simply <math>{14\choose5} = 2002</math>, because we just count how many ways we can place all <math>5</math> men in any of the <math>14</math> 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 | + | 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 <math>2</math> spots. |
There are <math>14</math> possible slots the first man can be placed. Once that man was placed, the next man only have <math>12</math> 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 <math>10</math> possible slots. The fourth man has <math>8</math> possible slots, and the fifth man has <math>6</math> possible slots. | There are <math>14</math> possible slots the first man can be placed. Once that man was placed, the next man only have <math>12</math> 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 <math>10</math> possible slots. The fourth man has <math>8</math> possible slots, and the fifth man has <math>6</math> possible slots. | ||
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~ericshi1685 | ~ericshi1685 | ||
− | ==Solution | + | ==Solution 6== |
We will first assign seats to the men. The first man can be placed in any of the <math>14</math> slots. The second man can be placed in any of the remaining <math>13</math> seats, except for the one diametrically opposite to the first man. So, there are <math>13 - 1 = 12</math> ways to seat him. With a similar argument, the third man can be seated in <math>10</math> ways, the fourth man in <math>8</math> ways and the last man in <math>6</math> ways. | We will first assign seats to the men. The first man can be placed in any of the <math>14</math> slots. The second man can be placed in any of the remaining <math>13</math> seats, except for the one diametrically opposite to the first man. So, there are <math>13 - 1 = 12</math> ways to seat him. With a similar argument, the third man can be seated in <math>10</math> ways, the fourth man in <math>8</math> ways and the last man in <math>6</math> ways. | ||
− | So, the total number of ways to arrange the men is <math>14 \ | + | So, the total number of ways to arrange the men is <math>14 \cdot 12 \cdot 10 \cdot 8 \cdot 6</math>. |
− | |||
− | |||
− | |||
− | |||
− | The answer is <math>48 + 143 = \boxed{191}</math> | + | The women go to the remaining <math>9</math> 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 <math>9!</math>, meaning that the total number of ways to arrange the people with restrictions is <math>14 \cdot 12 \cdot 10 \cdot 8 \cdot 6 \cdot 9!</math> In general, there are <math>14!</math> ways to arrange the people without restrictions. So, the probability is |
+ | <cmath>\frac{14 \cdot 12 \cdot 10 \cdot 8 \cdot 6 \cdot 9!}{14!} = \frac{8 \cdot 6}{13 \cdot 11} = \frac{48}{143}.</cmath> | ||
+ | The answer is <math>48 + 143 = \boxed{191}</math>. | ||
~baassid24 | ~baassid24 | ||
− | ==Solution | + | ==Solution 7== |
First pin one man on one seat (to ensure no rotate situations). Then there are <math>13!</math> arrangements. | First pin one man on one seat (to ensure no rotate situations). Then there are <math>13!</math> arrangements. | ||
− | Because 5 men must have women at their opposite side, we consider the | + | Because <math>5</math> men must have women at their opposite side, we consider the <math>2</math>nd man and the woman opposite as one group and name it <math>P_2.</math> There are <math>4</math> groups, <math>P_1, P_2, P_3, P_4</math> except the first man pinned on the same point. And for the rest <math>4</math> women, name them <math>P_5</math> and <math>P_6.</math> First to order <math>P_1, P_2, P_3, P_4, P_5, P_6,</math> there are <math>6!</math> ways. For the <math>1</math>st man, there are <math>9</math> women to choose, <math>8</math> for the <math>2</math>nd, <math>\ldots,</math> <math>5</math> for the <math>5</math>th, and then for the <math>2</math> women pairs <math>3</math> and <math>1.</math> Because every <math>2</math> person in the group have chance to change their position, there are <math>2^6</math> possibilities. |
− | So the possibility is <cmath>P=\frac{6!\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 3\cdot 1 \cdot 2^6}{13!}=\frac{48}{143}</cmath> | + | So the possibility is <cmath>P=\frac{6!\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 3\cdot 1 \cdot 2^6}{13!}=\frac{48}{143}.</cmath> |
− | The answer is <math>48+143=\boxed{191}</math> | + | The answer is <math>48+143=\boxed{191}.</math> |
~PLASTA | ~PLASTA | ||
− | ==Solution | + | ==Solution 8== |
We get around the condition that each man can't be opposite to another man by simply considering all <math>7</math> diagonals, and choosing <math>5</math> where there will be a single man. For each diagonal, the man can go on either side, and there are <math>\binom{14}{5}</math> ways to arrange the men and the women in total. Thus our answer is <math>\frac{\binom{7}{5}\cdot 2^5}{\binom{14}{5}} = \frac{48}{143}.</math> | We get around the condition that each man can't be opposite to another man by simply considering all <math>7</math> diagonals, and choosing <math>5</math> where there will be a single man. For each diagonal, the man can go on either side, and there are <math>\binom{14}{5}</math> ways to arrange the men and the women in total. Thus our answer is <math>\frac{\binom{7}{5}\cdot 2^5}{\binom{14}{5}} = \frac{48}{143}.</math> | ||
We get <math>48 + 143 = \boxed{191}</math> | We get <math>48 + 143 = \boxed{191}</math> | ||
~AtharvNaphade | ~AtharvNaphade | ||
+ | |||
+ | ==Solution 9== | ||
+ | We can find the probability of one arrangement occurring, and multiply it by the total number of arrangements. | ||
+ | |||
+ | The probability of a man being in any specific position is <math>\frac{5}{14}.</math> The probability of a woman being across from him is <math>\frac{9}{13}.</math> The probability of a man being in any valid position is now <math>\frac{4}{12},</math> and the probability of a woman being across from him is <math>\frac{8}{11},</math> and so forth. We stop when there are no more men left. Multiplying these probabilities together, | ||
+ | |||
+ | <cmath>P(\mathrm{One\ successful\ outcome})=\frac{5}{14}\cdot \frac{9}{13}\cdot \frac{4}{12}\cdot \frac{8}{11}\cdot \frac{3}{10}\cdot \frac{7}{9}\cdot \frac{2}{8}\cdot \frac{1}{6}\cdot \frac{6}{7}\cdot \frac{5}{5} = \frac{1}{2002}.</cmath> | ||
+ | To find the total number of successful outcomes, we consider the diagonals; the total number of diagonals to be made is <math>\binom75</math>, since there are <math>7</math> total diagonals, and we want to choose <math>5</math> of them to connect a man to a woman. For each of these diagonals, the man can be on either side of the diagonal. | ||
+ | |||
+ | It follows that there are <math>2</math> possibilities for each diagonal (man on one side, woman on the other, and vice versa). There are <math>5</math> diagonals with a man and a woman, so there are <math>2^5</math> different ways for these diagonals to appear. | ||
+ | |||
+ | There are <math>\binom75</math> successful diagonals, and for each of these diagonals, there are <math>2^5</math> ways to seat the men and the women, there are <math>\binom75</math> <math>\cdot 2^5</math> successful outcomes. | ||
+ | |||
+ | Recall that | ||
+ | <cmath>P(\mathrm{Any\ successful\ outcome})=P(\mathrm{One\ successful\ outcome})\cdot P(\mathrm{Total\ number\ of\ successful\ outcomes}).</cmath> | ||
+ | Therefore, | ||
+ | <cmath>P(\mathrm{Any\ successful\ outcome}) = \frac{1}{2002}\cdot 2^5\cdot \binom75 = \frac{1}{2002}\cdot 2^5\cdot 21 = \frac{2^5\cdot 21}{2002} = \frac{48}{143}.</cmath> | ||
+ | The requested sum is <math>48+143=\boxed{191}.</math> | ||
+ | |||
+ | -Benedict T (countmath1) | ||
+ | |||
+ | ==Video Solution (Mathematical Dexterity)== | ||
+ | |||
+ | https://www.youtube.com/watch?v=KdKysmdgepI | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/u7tG8P3vi5o | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
==See also== | ==See also== | ||
{{AIME box|year=2023|before=First Problem|num-a=2|n=I}} | {{AIME box|year=2023|before=First Problem|num-a=2|n=I}} | ||
− | |||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:17, 12 January 2024
Contents
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 (Inspired by Math Jam)
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 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
men must have women at their opposite side, we consider the
nd man and the woman opposite as one group and name it
There are
groups,
except the first man pinned on the same point. And for the rest
women, name them
and
First to order
there are
ways. For the
st man, there are
women to choose,
for the
nd,
for the
th, and then for the
women pairs
and
Because every
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
Solution 9
We can find the probability of one arrangement occurring, and multiply it by the total number of arrangements.
The probability of a man being in any specific position is The probability of a woman being across from him is
The probability of a man being in any valid position is now
and the probability of a woman being across from him is
and so forth. We stop when there are no more men left. Multiplying these probabilities together,
To find the total number of successful outcomes, we consider the diagonals; the total number of diagonals to be made is
, since there are
total diagonals, and we want to choose
of them to connect a man to a woman. For each of these diagonals, the man can be on either side of the diagonal.
It follows that there are possibilities for each diagonal (man on one side, woman on the other, and vice versa). There are
diagonals with a man and a woman, so there are
different ways for these diagonals to appear.
There are successful diagonals, and for each of these diagonals, there are
ways to seat the men and the women, there are
successful outcomes.
Recall that
Therefore,
The requested sum is
-Benedict T (countmath1)
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=KdKysmdgepI
Video Solution by TheBeautyofMath
~IceMatrix
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 | ||
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