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22 Mar 2004, 07:25
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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22 Mar 2004, 07:38
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)!
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hi [#permalink]

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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.
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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.
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22 Mar 2004, 11:42
C for me. I'll give details in a moment. Actually same as Geethu's reasoning!
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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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29 Mar 2004, 09:20
Not brag or anything but did I say B, I was right! haha.
[#permalink] 29 Mar 2004, 09:20
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