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

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Joined: 13 Nov 2010
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How many ways are there to split a group of 6 boys into two  [#permalink]

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Updated on: 09 Jul 2013, 10:15
13
00:00

Difficulty:

65% (hard)

Question Stats:

46% (00:34) correct 54% (00:48) wrong based on 292 sessions

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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

Originally posted by yufenshi on 25 Nov 2010, 17:06.
Last edited by Bunuel on 09 Jul 2013, 10:15, edited 1 time in total.
Edited the question.
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Joined: 02 Sep 2009
Posts: 49968
Re: ways to split a group of 6 boys into two groups of 3 boys ea  [#permalink]

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25 Nov 2010, 17:13
1
16
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!.

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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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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27 Nov 2010, 13:24
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]

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

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27 Nov 2010, 14:12
1
VeritasPrepKarishma wrote:
144144 wrote:
if i used the
6!/3!*(6-3)! * 1/2 is that ok as well?

Absolutely! You are doing 6C3 * (1/2).

I'd say it depends on the logic behind this formula.
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Re: How many ways are there to split a group of 6 boys into two  [#permalink]

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

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25 Sep 2018, 04:22
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Re: How many ways are there to split a group of 6 boys into two &nbs [#permalink] 25 Sep 2018, 04:22
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