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gurpreetsingh
Joined: 12 Oct 2009
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27 Sep 2010, 14:15
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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
A. 8
B. 10
C. 16
D. 20
E. 24
B
Show SpoilerDoubt
Why the answer is not 20? If we select 3 boys out of 6 , we can place them in one group and second group will be automatically selected.
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27 Sep 2010, 14:30
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gurpreetsingh
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 discussed at: combinations-problems-95344.html?hilit=dividing%20objects%20order#p734396
Hope it helps.
gurpreetsingh
Joined: 12 Oct 2009
Last visit: 15 Jun 2019
Posts: 2,272
Own Kudos:
Given Kudos: 235
Status:<strong>Nothing comes easy: neither do I want.</strong>
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Products:
27 Sep 2010, 14:52
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thanks I got it 
