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# http://www.manhattangmat.com

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Senior Manager
Joined: 05 May 2003
Posts: 427
Location: Aus
Followers: 2

Kudos [?]: 5 [0], given: 0

http://www.manhattangmat.com [#permalink]  22 Mar 2004, 07:25
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Senior Manager
Joined: 05 May 2003
Posts: 427
Location: Aus
Followers: 2

Kudos [?]: 5 [0], given: 0

C

n male and n female candidates have been recommended.
n/2 male and n/2 female candidates has to be selected from this lot.

using nCr formula we get

nC(n/2) * nC(n/2)

n! / (.5n)!(n-.5n)! * n! / (.5n)!(n-.5n)!
Eternal Intern
Joined: 22 Sep 2003
Posts: 51
Location: Illinois State University, Normal, IL
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hi [#permalink]  22 Mar 2004, 09:04
I guess B. I think C is wrong, in addition it was the answer last week.

I don't have any logic since this problem is to complicated.

But I think C is wrong cuz you can't pair the events together.
Eternal Intern
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Location: Illinois State University, Normal, IL
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hi [#permalink]  22 Mar 2004, 09:09
Yes it's B, using Pluggin IN Method

We can state that there is 4 offices,

That there is 2 men and 2 women

Combination 4,2 = 6

Using Choice B.

4! 4 x 3 x 2 x 1= 24

.5(4) =2 squared =4

24/4=6

so it's B.
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4312
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C for me. I'll give details in a moment. Actually same as Geethu's reasoning!
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Paul

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Re: hi [#permalink]  22 Mar 2004, 16:46
nikerun21 wrote:
Yes it's B, using Pluggin IN Method

We can state that there is 4 offices,

That there is 2 men and 2 women

Combination 4,2 = 6

Using Choice B.

4! 4 x 3 x 2 x 1= 24

.5(4) =2 squared =4

24/4=6

so it's B.

Nikerun,

I think in 4 offices we have 8 people, four men and four women.
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Eternal Intern
Joined: 22 Sep 2003
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Not brag or anything but did I say B, I was right! haha.
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