In how many different ways can a group of 8 people be

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In how many different ways can a group of 8 people be [#permalink]

13 Aug 2010, 08:38

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In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

A. 90
B. 105
C. 168
D. 420
E. 2520

[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Feb 2012, 15:05, edited 2 times in total.

Edited the question and added the OA

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Re: ways to divide?? [#permalink]

13 Aug 2010, 08:59

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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.

You can check similar problems:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+groups

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.
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Re: In how many different ways can a group of 8 people be [#permalink]

24 Mar 2012, 12:13

Another way to think about this:

How many ways can you arrange the following:
T1 T1 T2 T2 T3 T3 T4 T4

That would be: 8!/(2!*2!*2!*2!)

Then also recall that we don't care about the differences between the teams, therefore

8!/(2!*2!*2!*2!*4!) = 105

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Re: ways to divide?? [#permalink]

04 Sep 2012, 15:45

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)?

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Re: In how many different ways can a group of 8 people be [#permalink]

04 Sep 2012, 22:35

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We don't worry about the order within a team or between teams, because there is nothing in the question that would make it necessary to do so.

The question needs to give us information, why we should differentiate. For example, if those teams are given numbers to or being seated on a bench.

Or if one of the team members is the captain.

But that's not the case here.

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Re: ways to divide?? [#permalink]

05 Sep 2012, 00:39

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.

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.

Hope it helps.
_________________

Re: ways to divide?? [#permalink] 05 Sep 2012, 00:39

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