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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] New post 25 Nov 2010, 16:06
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A
B
C
D
E

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Question Stats:

46% (01:25) correct 54% (00:29) wrong based on 67 sessions
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, 09:15, 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] New post 25 Nov 2010, 16:13
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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] New post 25 Nov 2010, 20:49
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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] New post 27 Nov 2010, 12:24
if i used the
6!/3!*(6-3)! * 1/2 is that ok as well?
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Re: ways to split a group of 6 boys into two groups of 3 boys ea [#permalink] New post 27 Nov 2010, 12:44
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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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Re: ways to split a group of 6 boys into two groups of 3 boys ea [#permalink] New post 27 Nov 2010, 13:12
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Re: How many ways are there to split a group of 6 boys into two [#permalink] New post 17 Aug 2014, 03:17
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Re: How many ways are there to split a group of 6 boys into two   [#permalink] 17 Aug 2014, 03:17
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