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How many ways are there to split a group of 6 boys into two

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How many ways are there to split a group of 6 boys into two [#permalink]

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New post 25 Nov 2010, 17:06
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How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)

A. 8
B. 10
C. 16
D. 20
E. 24
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Jul 2013, 10:15, edited 1 time in total.
Edited the question.

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Re: ways to split a group of 6 boys into two groups of 3 boys ea [#permalink]

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New post 25 Nov 2010, 17:13
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yufenshi wrote:
How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)


8
10
16
20
24

The official answer is B, but I don't understand why in this case we need to divide by 2. It seems to me to be the same questions as: how many ways to choose 3 people out of 6 people. in that case it would be 20.

Thanks!


GENERAL RULE:
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 important is \(\frac{(mn)!}{(n!)^m}\)

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 NOT important is \(\frac{(mn)!}{(n!)^m*m!}\).

BACK TO THE ORIGINAL QUESTION:
In original question as the order is NOT important, we should use second formula, \(mn=6\), \(m=2\) groups \(n=3\) objects (people):
\(\frac{(mn)!}{(n!)^m*m!}=\frac{6!}{(3!)^2*2!}=10\).

This can be done in another way as well: \(\frac{C^3_6*C^3_3}{2!}=10\), we are dividing by \(2!\) as there are 2 groups and order doesn't matter.

For example if we choose with \(C^3_6\) the group {ABC} then the group {DEF} is left and we have two groups {ABC} and {DEF} but then we could choose also {DEF}, so in this case second group would be {ABC}, so we would have the same two groups: {ABC} and {DEF}. So to get rid of such duplications we should divide \(C^3_6*C^3_3\) by factorial of number of groups - 2!.

Answer: B.

This concept is also discussed at:
combinations-problems-95344.html?hilit=dividing%20objects%20order#p734396
split-the-group-101813.html?hilit=split
9-people-and-combinatorics-101722.html?hilit=divided%20equally%20into#p788744
ways-to-divide-99053.html?hilit=divided%20equally%20into#p763471

Hope it helps.
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Re: ways to split a group of 6 boys into two groups of 3 boys ea [#permalink]

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New post 25 Nov 2010, 21:49
and after going through Bunuel's explanation, to ensure that you have understood the concept, try the question in this post: http://gmatclub.com/forum/combination-105384.html

It the same reason why you do not multiply by 2 in the question in this post.
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Re: ways to split a group of 6 boys into two groups of 3 boys ea [#permalink]

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New post 27 Nov 2010, 13:24
if i used the
6!/3!*(6-3)! * 1/2 is that ok as well?
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New post 27 Nov 2010, 13:44
144144 wrote:
if i used the
6!/3!*(6-3)! * 1/2 is that ok as well?


Absolutely! You are doing 6C3 * (1/2).
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New post 26 Nov 2015, 16:40
Or simply put it in an anagram grid as: BBB|BBB , so we have 6!/3!*3!*2!=6*5*4/3*2*2=10.

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