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# simple cominatorics problem... i think oa is wrong

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Senior Manager
Joined: 10 Dec 2008
Posts: 476

Kudos [?]: 257 [0], given: 12

Location: United States
GMAT 1: 760 Q49 V44
GPA: 3.9
simple cominatorics problem... i think oa is wrong [#permalink]

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06 Aug 2009, 09:56
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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)

(C) 2008 GMAT Club - Probability and Combinations#16

* 8
* 10
* 16
* 20
* 24

[Reveal] Spoiler:
I think the answer is d, but the OA says:

Because the groups are not ordered, we must use formula $$\frac{C_6^3}{2} = \frac{\frac{6!}{3!3!}}{2} = 10$$ . We have to divide $$C_6^3$$ by 2 because we need to select one group of 3 boys and the other group will be automatically selected. In other words, the three boys left after selecting of 3 from 6 will constitute the other group of 3. That's why we need to consider only the half of instances of $$C_6^3$$ .

Kudos [?]: 257 [0], given: 12

Senior Manager
Joined: 17 Jul 2009
Posts: 285

Kudos [?]: 44 [0], given: 9

Concentration: Nonprofit, Strategy
GPA: 3.42
WE: Engineering (Computer Hardware)
Re: simple cominatorics problem... i think oa is wrong [#permalink]

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06 Aug 2009, 12:34
order of group doesn't matter as if you choose A B C for the first group, then it's the same as choosing DEF for the 1st group....thus the (6C3)/2..great find.

Kudos [?]: 44 [0], given: 9

Senior Manager
Joined: 10 Dec 2008
Posts: 476

Kudos [?]: 257 [0], given: 12

Location: United States
GMAT 1: 760 Q49 V44
GPA: 3.9
Re: simple cominatorics problem... i think oa is wrong [#permalink]

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06 Aug 2009, 13:22
i see... tricky

Kudos [?]: 257 [0], given: 12

Re: simple cominatorics problem... i think oa is wrong   [#permalink] 06 Aug 2009, 13:22
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