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Re: ways to divide?? [#permalink]
13 Aug 2010, 07:59

8

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

bibha wrote:

In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? 90 105 168 420 2520

\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way. For the first person we can pick a pair in 7 ways; For the second one in 5 ways (as two are already chosen); For the third one in 3 ways (as 4 people are already chosen); For the fourth one there is only one left.

1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \frac{(mn)!}{(n!)^m*m!}.

2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is \frac{(mn)!}{(n!)^m}

Re: ways to divide?? [#permalink]
04 Sep 2012, 14:45

Bunuel wrote:

bibha wrote:

In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? 90 105 168 420 2520

\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way. For the first person we can pick a pair in 7 ways; For the second one in 5 ways (as two are already chosen); For the third one in 3 ways (as 4 people are already chosen); For the fourth one there is only one left.

So we have 7*5*3*1=105

Answer: B. There is also direct formula for this:

1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \frac{(mn)!}{(n!)^m*m!}.

2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is \frac{(mn)!}{(n!)^m}

Hope it helps.

I understand how to get the answer, but I'm wondering why we assume that the order of the teams doesn't matter. The question just asks how many ways you can group them. When should we assume something matters or doesn't matter when the question doesn't specify? And also, just to clarify, when we do 8!/2!2!2!2!4!, we are also not worrying about the order within each team either right i.e. (a,b)=(b,a)?

Re: ways to divide?? [#permalink]
04 Sep 2012, 23:39

1

This post received KUDOS

Expert's post

dandarth1 wrote:

Bunuel wrote:

bibha wrote:

In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? 90 105 168 420 2520

\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way. For the first person we can pick a pair in 7 ways; For the second one in 5 ways (as two are already chosen); For the third one in 3 ways (as 4 people are already chosen); For the fourth one there is only one left.

So we have 7*5*3*1=105

Answer: B. There is also direct formula for this:

1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \frac{(mn)!}{(n!)^m*m!}.

2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is \frac{(mn)!}{(n!)^m}

Hope it helps.

I understand how to get the answer, but I'm wondering why we assume that the order of the teams doesn't matter. The question just asks how many ways you can group them. When should we assume something matters or doesn't matter when the question doesn't specify? And also, just to clarify, when we do 8!/2!2!2!2!4!, we are also not worrying about the order within each team either right i.e. (a,b)=(b,a)?

The teams are not numbered/labeled (we don't have team #1, #2, ...), the teams are not assigned to something (for example to tournaments), ... So, the order of the teams doesn't matter.

Please check the links in my previous post for similar problems.

Re: In how many different ways can a group of 8 people be [#permalink]
15 Sep 2013, 02:25

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Re: In how many different ways can a group of 8 people be [#permalink]
08 Nov 2014, 02:13

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