How many ways are there to split $10$ people into two groups of $5$? [duplicate]

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This question already has an answer here:

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  How many ways can N elements be partitioned into subsets of size K?
  
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In the example below I get the first part. May I know why they're dividing by 2 for the second part? I'm asking because I feel the answer for second part should also be $binom105$.

In the first part simply by choosing 5 people, we're already splitting the squad into two teams. So I don't really see a difference between parts 1 & 2. Help?

Example

There are ten people in a basketball squad. Find how many ways:

1. the starting five can be chosen from the squad


2. the squad can be split into two teams of five.

Solution

1. There are $binom105 = frac10times9times8times7times65times4times3times2times1 = 252$ ways of chosing the starting five.


2. The number of ways of dividing the squad into two teams of five is $frac2522 = 126$.

combinatorics

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edited Jul 20 at 3:15

Xander Henderson

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asked Jul 19 at 12:35

rsadhvika

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marked as duplicate by Xander Henderson, Rhys Steele, Simply Beautiful Art, Did, Parcly Taxel Jul 20 at 13:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  10
  

  
  
  
  
  
  

  
  in the first case, the order of team does matter, in the second one it does not
  – Vasya
  Jul 19 at 13:00
  

  
  
  
  

add a comment | 

up vote
14
down vote

favorite

5

This question already has an answer here:

• 
  How many ways can N elements be partitioned into subsets of size K?
  
  2 answers
  
  

In the example below I get the first part. May I know why they're dividing by 2 for the second part? I'm asking because I feel the answer for second part should also be $binom105$.

In the first part simply by choosing 5 people, we're already splitting the squad into two teams. So I don't really see a difference between parts 1 & 2. Help?

Example

There are ten people in a basketball squad. Find how many ways:

1. the starting five can be chosen from the squad


2. the squad can be split into two teams of five.

Solution

1. There are $binom105 = frac10times9times8times7times65times4times3times2times1 = 252$ ways of chosing the starting five.


2. The number of ways of dividing the squad into two teams of five is $frac2522 = 126$.

combinatorics

share|cite|improve this question

edited Jul 20 at 3:15

Xander Henderson

13.1k83150

asked Jul 19 at 12:35

rsadhvika

1,4891026

marked as duplicate by Xander Henderson, Rhys Steele, Simply Beautiful Art, Did, Parcly Taxel Jul 20 at 13:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  10
  

  
  
  
  
  
  

  
  in the first case, the order of team does matter, in the second one it does not
  – Vasya
  Jul 19 at 13:00
  

  
  
  
  

add a comment | 

up vote
14
down vote

favorite

5

up vote
14
down vote

favorite

5

5

This question already has an answer here:

• 
  How many ways can N elements be partitioned into subsets of size K?
  
  2 answers
  
  

In the example below I get the first part. May I know why they're dividing by 2 for the second part? I'm asking because I feel the answer for second part should also be $binom105$.

In the first part simply by choosing 5 people, we're already splitting the squad into two teams. So I don't really see a difference between parts 1 & 2. Help?

Example

There are ten people in a basketball squad. Find how many ways:

1. the starting five can be chosen from the squad


2. the squad can be split into two teams of five.

Solution

1. There are $binom105 = frac10times9times8times7times65times4times3times2times1 = 252$ ways of chosing the starting five.


2. The number of ways of dividing the squad into two teams of five is $frac2522 = 126$.

combinatorics

share|cite|improve this question

edited Jul 20 at 3:15

Xander Henderson

13.1k83150

asked Jul 19 at 12:35

rsadhvika

1,4891026

This question already has an answer here:

• 
  How many ways can N elements be partitioned into subsets of size K?
  
  2 answers
  
  

In the example below I get the first part. May I know why they're dividing by 2 for the second part? I'm asking because I feel the answer for second part should also be $binom105$.

In the first part simply by choosing 5 people, we're already splitting the squad into two teams. So I don't really see a difference between parts 1 & 2. Help?

Example

There are ten people in a basketball squad. Find how many ways:

1. the starting five can be chosen from the squad


2. the squad can be split into two teams of five.

Solution

1. There are $binom105 = frac10times9times8times7times65times4times3times2times1 = 252$ ways of chosing the starting five.


2. The number of ways of dividing the squad into two teams of five is $frac2522 = 126$.

This question already has an answer here:

• 
  How many ways can N elements be partitioned into subsets of size K?
  
  2 answers
  
  

combinatorics

share|cite|improve this question

edited Jul 20 at 3:15

Xander Henderson

13.1k83150

asked Jul 19 at 12:35

rsadhvika

1,4891026

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 3:15

Xander Henderson

13.1k83150

edited Jul 20 at 3:15

Xander Henderson

13.1k83150

edited Jul 20 at 3:15

Xander Henderson

13.1k83150

13.1k83150

asked Jul 19 at 12:35

rsadhvika

1,4891026

asked Jul 19 at 12:35

rsadhvika

1,4891026

asked Jul 19 at 12:35

rsadhvika

1,4891026

1,4891026

marked as duplicate by Xander Henderson, Rhys Steele, Simply Beautiful Art, Did, Parcly Taxel Jul 20 at 13:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Xander Henderson, Rhys Steele, Simply Beautiful Art, Did, Parcly Taxel Jul 20 at 13:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  10
  

  
  
  
  
  
  

  
  in the first case, the order of team does matter, in the second one it does not
  – Vasya
  Jul 19 at 13:00
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  10
  

  
  
  
  
  
  

  
  in the first case, the order of team does matter, in the second one it does not
  – Vasya
  Jul 19 at 13:00
  

  
  
  
  

10

10

in the first case, the order of team does matter, in the second one it does not
– Vasya
Jul 19 at 13:00

in the first case, the order of team does matter, in the second one it does not
– Vasya
Jul 19 at 13:00

add a comment | 

7 Answers
7

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oldest

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up vote
13
down vote



accepted

In the first one, there is a difference between the two resulting teams. One is playing, the other is sitting on the sidelines. So swapping the two teams around makes for a different choice of starting five.

In the second case, there is no difference. The only thing that matters is who is on team with who, and not which team is team A and what team is team B. So swapping the two teams around still gives the same choice of two teams.

You can also go the other way: Choose a starting five by first dividing into two teams (which can be done in some number of ways), then choosing which of the two teams gets to be athe starting five. You have two options for starting five for each division into two teams. Thus you have to multiply by 2 to get from one to the other.

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edited Jul 19 at 12:43

answered Jul 19 at 12:38

Arthur

98.8k793175

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up vote
7
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Observe that the choice of choosing five people and the other five are counted separately in the first case, but yield the same separation into two groups in the second. That's why there is division by $2$.

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answered Jul 19 at 12:37

uniquesolution

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up vote
7
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Note that when we choose $binom105$ we are counting twice the same teams, as for example

$$ A,B,C,D,E , G,H,I,L,M equiv G,H,I,L,M ,A,B,C,D,E$$

therefore the number of ways to obtain 2 teams is $frac12 binom105$.

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answered Jul 19 at 12:45

gimusi

65.4k73584

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up vote
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In the first case each choice is about selecting 5 out of ten, let's say from $A,B,C,D,E,F,G,H,I,J$. Note that selecting $A,B,C,D,E$ id different, than selecting $F,G,H,I,J$.

In the second case you select 5, but the other 5 are also forming a team. Thus your selection is for example $A,B,C,D,E,F,G,H,I,J$. This selection is the same, as selecting $F,G,H,I,J,A,B,C,D,E$. Thus in the way, that you form two teams by selecting 5 people to one and rest to another leads to duplication of selections. Therefore you need to divide it by the number of permutations of the teams, ie $2!$.

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answered Jul 19 at 12:47

Jaroslaw Matlak

3,880830

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up vote
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Imagine that you're one of the $10$ players. In both cases you need to determine who your $4$ teammates will be, out of the other $9$ players. This can be done in $binom94 = 126$ ways.

Now, if the two teams are treated differently, like one team gets to play while the other team has to sit on the bench, then (once you have determined your teammates) you still have to determine which of the $2$ possible assignments your team gets. The two outcomes are rather different for you, so we have to count them differently. There are thus $2binom94 = 252$ possible outcomes.

On the other hand, if there is no difference between teams, like if the two teams are just going to play each other, then they are not treated differently. It is not really possible to distinguish between the following cases:

(I) You, Alice, Bob, Charlie and Danielle are going to play against the other five players

or

(II) The other five players are going to play against you, Alice, Bob, Charlie and Danielle.

Both (I) and (II) describe the same event. That event should not be counted twice. So there are just $binom94 = 126$ possible outcomes here.

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answered Jul 19 at 23:39

mathmandan

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up vote
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This is from Sheldon Ross's book a First Course in Probability, I remember that example confusing me. Sometimes it helped me to think of it this way - assume that the first 5 picked will be on the red jersey team and the remaining will be on the yellow jersey team for option 1. Then a team of A,B,C,D,E in red jerseys is different than a team A,B,C,D,E in yellow jerseys, so they are both counted. In option 2, this is irrelevant so we divide by 2 to take out this distinction.

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answered Jul 19 at 16:21

Scott Hootman-Ng

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The second part of your problem will be $binom105$ if the question is of the following form:

How many ways to split $10$ people into two groups of $5$ people so that the first group is Team A and second group is Team B?

Here, unlike the original question, we restrict ourselves to a particular arrangement between the groups. If we don't have such a restriction as in the original post, we will have to divide by the number of possibilities to order the groups which is $2!$. Then the answer would become
$$fracbinom1052!=126.$$

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answered Jul 19 at 16:21

Bodhi

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7 Answers
7

active

oldest

votes

7 Answers
7

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
13
down vote



accepted

In the first one, there is a difference between the two resulting teams. One is playing, the other is sitting on the sidelines. So swapping the two teams around makes for a different choice of starting five.

In the second case, there is no difference. The only thing that matters is who is on team with who, and not which team is team A and what team is team B. So swapping the two teams around still gives the same choice of two teams.

You can also go the other way: Choose a starting five by first dividing into two teams (which can be done in some number of ways), then choosing which of the two teams gets to be athe starting five. You have two options for starting five for each division into two teams. Thus you have to multiply by 2 to get from one to the other.

share|cite|improve this answer

edited Jul 19 at 12:43

answered Jul 19 at 12:38

Arthur

98.8k793175

add a comment | 

up vote
13
down vote



accepted

In the first one, there is a difference between the two resulting teams. One is playing, the other is sitting on the sidelines. So swapping the two teams around makes for a different choice of starting five.

In the second case, there is no difference. The only thing that matters is who is on team with who, and not which team is team A and what team is team B. So swapping the two teams around still gives the same choice of two teams.

You can also go the other way: Choose a starting five by first dividing into two teams (which can be done in some number of ways), then choosing which of the two teams gets to be athe starting five. You have two options for starting five for each division into two teams. Thus you have to multiply by 2 to get from one to the other.

share|cite|improve this answer

edited Jul 19 at 12:43

answered Jul 19 at 12:38

Arthur

98.8k793175

add a comment | 

up vote
13
down vote



accepted

up vote
13
down vote



accepted

In the first one, there is a difference between the two resulting teams. One is playing, the other is sitting on the sidelines. So swapping the two teams around makes for a different choice of starting five.

In the second case, there is no difference. The only thing that matters is who is on team with who, and not which team is team A and what team is team B. So swapping the two teams around still gives the same choice of two teams.

You can also go the other way: Choose a starting five by first dividing into two teams (which can be done in some number of ways), then choosing which of the two teams gets to be athe starting five. You have two options for starting five for each division into two teams. Thus you have to multiply by 2 to get from one to the other.

share|cite|improve this answer

edited Jul 19 at 12:43

answered Jul 19 at 12:38

Arthur

98.8k793175

In the first one, there is a difference between the two resulting teams. One is playing, the other is sitting on the sidelines. So swapping the two teams around makes for a different choice of starting five.

In the second case, there is no difference. The only thing that matters is who is on team with who, and not which team is team A and what team is team B. So swapping the two teams around still gives the same choice of two teams.

You can also go the other way: Choose a starting five by first dividing into two teams (which can be done in some number of ways), then choosing which of the two teams gets to be athe starting five. You have two options for starting five for each division into two teams. Thus you have to multiply by 2 to get from one to the other.

share|cite|improve this answer

edited Jul 19 at 12:43

answered Jul 19 at 12:38

Arthur

98.8k793175

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 12:43

edited Jul 19 at 12:43

edited Jul 19 at 12:43

answered Jul 19 at 12:38

Arthur

98.8k793175

answered Jul 19 at 12:38

Arthur

98.8k793175

answered Jul 19 at 12:38

Arthur

98.8k793175

98.8k793175

add a comment | 

add a comment | 

up vote
7
down vote

Observe that the choice of choosing five people and the other five are counted separately in the first case, but yield the same separation into two groups in the second. That's why there is division by $2$.

share|cite|improve this answer

answered Jul 19 at 12:37

uniquesolution

7,700721

add a comment | 

up vote
7
down vote

Observe that the choice of choosing five people and the other five are counted separately in the first case, but yield the same separation into two groups in the second. That's why there is division by $2$.

share|cite|improve this answer

answered Jul 19 at 12:37

uniquesolution

7,700721

add a comment | 

up vote
7
down vote

up vote
7
down vote

Observe that the choice of choosing five people and the other five are counted separately in the first case, but yield the same separation into two groups in the second. That's why there is division by $2$.

share|cite|improve this answer

answered Jul 19 at 12:37

uniquesolution

7,700721

Observe that the choice of choosing five people and the other five are counted separately in the first case, but yield the same separation into two groups in the second. That's why there is division by $2$.

share|cite|improve this answer

answered Jul 19 at 12:37

uniquesolution

7,700721

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 12:37

uniquesolution

7,700721

answered Jul 19 at 12:37

uniquesolution

7,700721

answered Jul 19 at 12:37

uniquesolution

7,700721

7,700721

add a comment | 

add a comment | 

up vote
7
down vote

Note that when we choose $binom105$ we are counting twice the same teams, as for example

$$ A,B,C,D,E , G,H,I,L,M equiv G,H,I,L,M ,A,B,C,D,E$$

therefore the number of ways to obtain 2 teams is $frac12 binom105$.

share|cite|improve this answer

answered Jul 19 at 12:45

gimusi

65.4k73584

add a comment | 

up vote
7
down vote

Note that when we choose $binom105$ we are counting twice the same teams, as for example

$$ A,B,C,D,E , G,H,I,L,M equiv G,H,I,L,M ,A,B,C,D,E$$

therefore the number of ways to obtain 2 teams is $frac12 binom105$.

share|cite|improve this answer

answered Jul 19 at 12:45

gimusi

65.4k73584

add a comment | 

up vote
7
down vote

up vote
7
down vote

Note that when we choose $binom105$ we are counting twice the same teams, as for example

$$ A,B,C,D,E , G,H,I,L,M equiv G,H,I,L,M ,A,B,C,D,E$$

therefore the number of ways to obtain 2 teams is $frac12 binom105$.

share|cite|improve this answer

answered Jul 19 at 12:45

gimusi

65.4k73584

Note that when we choose $binom105$ we are counting twice the same teams, as for example

$$ A,B,C,D,E , G,H,I,L,M equiv G,H,I,L,M ,A,B,C,D,E$$

therefore the number of ways to obtain 2 teams is $frac12 binom105$.

share|cite|improve this answer

answered Jul 19 at 12:45

gimusi

65.4k73584

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 12:45

gimusi

65.4k73584

answered Jul 19 at 12:45

gimusi

65.4k73584

answered Jul 19 at 12:45

gimusi

65.4k73584

65.4k73584

add a comment | 

add a comment | 

up vote
5
down vote

In the first case each choice is about selecting 5 out of ten, let's say from $A,B,C,D,E,F,G,H,I,J$. Note that selecting $A,B,C,D,E$ id different, than selecting $F,G,H,I,J$.

In the second case you select 5, but the other 5 are also forming a team. Thus your selection is for example $A,B,C,D,E,F,G,H,I,J$. This selection is the same, as selecting $F,G,H,I,J,A,B,C,D,E$. Thus in the way, that you form two teams by selecting 5 people to one and rest to another leads to duplication of selections. Therefore you need to divide it by the number of permutations of the teams, ie $2!$.

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answered Jul 19 at 12:47

Jaroslaw Matlak

3,880830

add a comment | 

up vote
5
down vote

In the first case each choice is about selecting 5 out of ten, let's say from $A,B,C,D,E,F,G,H,I,J$. Note that selecting $A,B,C,D,E$ id different, than selecting $F,G,H,I,J$.

In the second case you select 5, but the other 5 are also forming a team. Thus your selection is for example $A,B,C,D,E,F,G,H,I,J$. This selection is the same, as selecting $F,G,H,I,J,A,B,C,D,E$. Thus in the way, that you form two teams by selecting 5 people to one and rest to another leads to duplication of selections. Therefore you need to divide it by the number of permutations of the teams, ie $2!$.

share|cite|improve this answer

answered Jul 19 at 12:47

Jaroslaw Matlak

3,880830

add a comment | 

up vote
5
down vote

up vote
5
down vote

In the first case each choice is about selecting 5 out of ten, let's say from $A,B,C,D,E,F,G,H,I,J$. Note that selecting $A,B,C,D,E$ id different, than selecting $F,G,H,I,J$.

In the second case you select 5, but the other 5 are also forming a team. Thus your selection is for example $A,B,C,D,E,F,G,H,I,J$. This selection is the same, as selecting $F,G,H,I,J,A,B,C,D,E$. Thus in the way, that you form two teams by selecting 5 people to one and rest to another leads to duplication of selections. Therefore you need to divide it by the number of permutations of the teams, ie $2!$.

share|cite|improve this answer

answered Jul 19 at 12:47

Jaroslaw Matlak

3,880830

In the first case each choice is about selecting 5 out of ten, let's say from $A,B,C,D,E,F,G,H,I,J$. Note that selecting $A,B,C,D,E$ id different, than selecting $F,G,H,I,J$.

In the second case you select 5, but the other 5 are also forming a team. Thus your selection is for example $A,B,C,D,E,F,G,H,I,J$. This selection is the same, as selecting $F,G,H,I,J,A,B,C,D,E$. Thus in the way, that you form two teams by selecting 5 people to one and rest to another leads to duplication of selections. Therefore you need to divide it by the number of permutations of the teams, ie $2!$.

share|cite|improve this answer

answered Jul 19 at 12:47

Jaroslaw Matlak

3,880830

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 12:47

Jaroslaw Matlak

3,880830

answered Jul 19 at 12:47

Jaroslaw Matlak

3,880830

answered Jul 19 at 12:47

Jaroslaw Matlak

3,880830

3,880830

add a comment | 

add a comment | 

up vote
3
down vote

Imagine that you're one of the $10$ players. In both cases you need to determine who your $4$ teammates will be, out of the other $9$ players. This can be done in $binom94 = 126$ ways.

Now, if the two teams are treated differently, like one team gets to play while the other team has to sit on the bench, then (once you have determined your teammates) you still have to determine which of the $2$ possible assignments your team gets. The two outcomes are rather different for you, so we have to count them differently. There are thus $2binom94 = 252$ possible outcomes.

On the other hand, if there is no difference between teams, like if the two teams are just going to play each other, then they are not treated differently. It is not really possible to distinguish between the following cases:

(I) You, Alice, Bob, Charlie and Danielle are going to play against the other five players

or

(II) The other five players are going to play against you, Alice, Bob, Charlie and Danielle.

Both (I) and (II) describe the same event. That event should not be counted twice. So there are just $binom94 = 126$ possible outcomes here.

share|cite|improve this answer

answered Jul 19 at 23:39

mathmandan

1,4251712

add a comment | 

up vote
3
down vote

Imagine that you're one of the $10$ players. In both cases you need to determine who your $4$ teammates will be, out of the other $9$ players. This can be done in $binom94 = 126$ ways.

Now, if the two teams are treated differently, like one team gets to play while the other team has to sit on the bench, then (once you have determined your teammates) you still have to determine which of the $2$ possible assignments your team gets. The two outcomes are rather different for you, so we have to count them differently. There are thus $2binom94 = 252$ possible outcomes.

On the other hand, if there is no difference between teams, like if the two teams are just going to play each other, then they are not treated differently. It is not really possible to distinguish between the following cases:

(I) You, Alice, Bob, Charlie and Danielle are going to play against the other five players

or

(II) The other five players are going to play against you, Alice, Bob, Charlie and Danielle.

Both (I) and (II) describe the same event. That event should not be counted twice. So there are just $binom94 = 126$ possible outcomes here.

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answered Jul 19 at 23:39

mathmandan

1,4251712

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up vote
3
down vote

up vote
3
down vote

Imagine that you're one of the $10$ players. In both cases you need to determine who your $4$ teammates will be, out of the other $9$ players. This can be done in $binom94 = 126$ ways.

Now, if the two teams are treated differently, like one team gets to play while the other team has to sit on the bench, then (once you have determined your teammates) you still have to determine which of the $2$ possible assignments your team gets. The two outcomes are rather different for you, so we have to count them differently. There are thus $2binom94 = 252$ possible outcomes.

On the other hand, if there is no difference between teams, like if the two teams are just going to play each other, then they are not treated differently. It is not really possible to distinguish between the following cases:

(I) You, Alice, Bob, Charlie and Danielle are going to play against the other five players

or

(II) The other five players are going to play against you, Alice, Bob, Charlie and Danielle.

Both (I) and (II) describe the same event. That event should not be counted twice. So there are just $binom94 = 126$ possible outcomes here.

share|cite|improve this answer

answered Jul 19 at 23:39

mathmandan

1,4251712

Imagine that you're one of the $10$ players. In both cases you need to determine who your $4$ teammates will be, out of the other $9$ players. This can be done in $binom94 = 126$ ways.

Now, if the two teams are treated differently, like one team gets to play while the other team has to sit on the bench, then (once you have determined your teammates) you still have to determine which of the $2$ possible assignments your team gets. The two outcomes are rather different for you, so we have to count them differently. There are thus $2binom94 = 252$ possible outcomes.

On the other hand, if there is no difference between teams, like if the two teams are just going to play each other, then they are not treated differently. It is not really possible to distinguish between the following cases:

(I) You, Alice, Bob, Charlie and Danielle are going to play against the other five players

or

(II) The other five players are going to play against you, Alice, Bob, Charlie and Danielle.

Both (I) and (II) describe the same event. That event should not be counted twice. So there are just $binom94 = 126$ possible outcomes here.

share|cite|improve this answer

answered Jul 19 at 23:39

mathmandan

1,4251712

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share|cite|improve this answer

answered Jul 19 at 23:39

mathmandan

1,4251712

answered Jul 19 at 23:39

mathmandan

1,4251712

answered Jul 19 at 23:39

mathmandan

1,4251712

1,4251712

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This is from Sheldon Ross's book a First Course in Probability, I remember that example confusing me. Sometimes it helped me to think of it this way - assume that the first 5 picked will be on the red jersey team and the remaining will be on the yellow jersey team for option 1. Then a team of A,B,C,D,E in red jerseys is different than a team A,B,C,D,E in yellow jerseys, so they are both counted. In option 2, this is irrelevant so we divide by 2 to take out this distinction.

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answered Jul 19 at 16:21

Scott Hootman-Ng

5115

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up vote
2
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This is from Sheldon Ross's book a First Course in Probability, I remember that example confusing me. Sometimes it helped me to think of it this way - assume that the first 5 picked will be on the red jersey team and the remaining will be on the yellow jersey team for option 1. Then a team of A,B,C,D,E in red jerseys is different than a team A,B,C,D,E in yellow jerseys, so they are both counted. In option 2, this is irrelevant so we divide by 2 to take out this distinction.

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answered Jul 19 at 16:21

Scott Hootman-Ng

5115

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up vote
2
down vote

up vote
2
down vote

This is from Sheldon Ross's book a First Course in Probability, I remember that example confusing me. Sometimes it helped me to think of it this way - assume that the first 5 picked will be on the red jersey team and the remaining will be on the yellow jersey team for option 1. Then a team of A,B,C,D,E in red jerseys is different than a team A,B,C,D,E in yellow jerseys, so they are both counted. In option 2, this is irrelevant so we divide by 2 to take out this distinction.

share|cite|improve this answer

answered Jul 19 at 16:21

Scott Hootman-Ng

5115

This is from Sheldon Ross's book a First Course in Probability, I remember that example confusing me. Sometimes it helped me to think of it this way - assume that the first 5 picked will be on the red jersey team and the remaining will be on the yellow jersey team for option 1. Then a team of A,B,C,D,E in red jerseys is different than a team A,B,C,D,E in yellow jerseys, so they are both counted. In option 2, this is irrelevant so we divide by 2 to take out this distinction.

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answered Jul 19 at 16:21

Scott Hootman-Ng

5115

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 16:21

Scott Hootman-Ng

5115

answered Jul 19 at 16:21

Scott Hootman-Ng

5115

answered Jul 19 at 16:21

Scott Hootman-Ng

5115

5115

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add a comment | 

up vote
2
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The second part of your problem will be $binom105$ if the question is of the following form:

How many ways to split $10$ people into two groups of $5$ people so that the first group is Team A and second group is Team B?

Here, unlike the original question, we restrict ourselves to a particular arrangement between the groups. If we don't have such a restriction as in the original post, we will have to divide by the number of possibilities to order the groups which is $2!$. Then the answer would become
$$fracbinom1052!=126.$$

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answered Jul 19 at 16:21

Bodhi

234

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The second part of your problem will be $binom105$ if the question is of the following form:

How many ways to split $10$ people into two groups of $5$ people so that the first group is Team A and second group is Team B?

Here, unlike the original question, we restrict ourselves to a particular arrangement between the groups. If we don't have such a restriction as in the original post, we will have to divide by the number of possibilities to order the groups which is $2!$. Then the answer would become
$$fracbinom1052!=126.$$

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answered Jul 19 at 16:21

Bodhi

234

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up vote
2
down vote

up vote
2
down vote

The second part of your problem will be $binom105$ if the question is of the following form:

How many ways to split $10$ people into two groups of $5$ people so that the first group is Team A and second group is Team B?

Here, unlike the original question, we restrict ourselves to a particular arrangement between the groups. If we don't have such a restriction as in the original post, we will have to divide by the number of possibilities to order the groups which is $2!$. Then the answer would become
$$fracbinom1052!=126.$$

share|cite|improve this answer

answered Jul 19 at 16:21

Bodhi

234

The second part of your problem will be $binom105$ if the question is of the following form:

How many ways to split $10$ people into two groups of $5$ people so that the first group is Team A and second group is Team B?

Here, unlike the original question, we restrict ourselves to a particular arrangement between the groups. If we don't have such a restriction as in the original post, we will have to divide by the number of possibilities to order the groups which is $2!$. Then the answer would become
$$fracbinom1052!=126.$$

share|cite|improve this answer

answered Jul 19 at 16:21

Bodhi

234

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share|cite|improve this answer

answered Jul 19 at 16:21

Bodhi

234

answered Jul 19 at 16:21

Bodhi

234

answered Jul 19 at 16:21

Bodhi

234

234

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