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gmat prep

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Director
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gmat prep [#permalink]

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New post 10 Sep 2008, 20:59
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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SVP
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Re: gmat prep [#permalink]

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New post 10 Sep 2008, 21:21
bigtreezl wrote:
see attached



= 4 x 6c2 + 4c2 x 6c2 + 4c3
= 100
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Director
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Re: gmat prep [#permalink]

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New post 11 Sep 2008, 17:37
bigtreezl wrote:
see attached


Total number of ways to choose committee = 10C3 = 120
Total number of ways to choose committee with no French teachers = 6C3 = 20

Total number of ways to choose committee with at least one French teacher = 120 - 20 = 100

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Manager
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Re: gmat prep [#permalink]

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New post 11 Sep 2008, 17:44
4 x 6c2 + 4c2 x 6c1 + 4c3
= ( one french ) * (2 people rest 6) + ( 2 out of french ) * ( 1 out of 6 ) + all 3 out of 4
= 100

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Re: gmat prep [#permalink]

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New post 11 Sep 2008, 19:02
bigtreezl wrote:
see attached

If 3 selected out 0f 10 ,we get 10C2=120 total selections or commities possible.

If 3 selected out of 6 for not french commities => 6C3=20 commities

hence 120-20 =100 commities for atleast one french teacher in commities

IMO E
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Director
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Re: gmat prep [#permalink]

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New post 12 Sep 2008, 00:45
great explanation zoinnk!

OA is E

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Re: gmat prep   [#permalink] 12 Sep 2008, 00:45
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