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In how many different ways can a group of 6 boys be divided into two teams, with each team consisting of 3 boys ?
A. 8
B. 10
C. 16
D. 20
E. 24
A. 8
B. 10
C. 16
D. 20
E. 24
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25 Nov 2010, 17:13
Kudos
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yufenshi
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:
The answer to the above question is given by \(\frac{C^3_6*C^3_3}{2!}\), which equals 10. The division by \(2!\), the factorial of the number of teams, is performed because the order of the groups does not matter.
For example, let's consider a group of 6 boys named A, B, C, D, E, and F. One possible grouping using \(C^3_6*C^3_3\) is {ABC} as the first group and {DEF} as the second group, resulting in two groups: {ABC} and {DEF}. However, \(C^3_6*C^3_3\) will also yield the group {DEF} with {ABC} as the second group, resulting in the same two groups: {ABC} and {DEF}. To avoid counting duplicates, we divide \(C^3_6*C^3_3\) by the factorial of the number of teams, which is 2!.
Answer: B
This concept is also discussed at:
https://gmatclub.com/forum/combinations ... er#p734396
https://gmatclub.com/forum/split-the-gr ... ilit=split
https://gmatclub.com/forum/9-people-and ... to#p788744
https://gmatclub.com/forum/ways-to-divi ... to#p763471
Hope it helps.
06 Sep 2011, 21:19
Kudos
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reatsaint
Out of 6 boys, you can choose 3 in 6C3 ways and make the first group. The second group is of the remaining 3 boys. But when you do that, you have ordered the groups into first and second. The question mentions that the order of the groups does not matter. Hence, you need to divide your answer i.e. 6C3 by 2! to undo the ordering of the groups.
That is, 6C3 gives you 20 ways. This includes
G1 - ABC and G2 - DEF
G1 - DEF and G2 - ABC
as two different ways but it is actually just one way because there is no G1 and G2. In both the cases, the two groups are ABC and DEF
Therefore, answer will be 20/2! = 10
