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# A Committee of 6 is chosen from 8 men and 5 women, so as to

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Intern
Joined: 26 Mar 2006
Posts: 27

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A Committee of 6 is chosen from 8 men and 5 women, so as to [#permalink]

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16 Apr 2006, 19:08
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A Committee of 6 is chosen from 8 men and 5 women, so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

A- 3510
B- 2620
C- 1404
D- 700
E- 635

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Senior Manager
Joined: 08 Sep 2004
Posts: 257

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Location: New York City, USA

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16 Apr 2006, 21:10
IMO E.

Thanks,
Vipin

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Intern
Joined: 17 Apr 2006
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17 Apr 2006, 21:48
can you explain how you got this number?

thanks,

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Manager
Joined: 28 Mar 2006
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18 Apr 2006, 09:37
1st case:-(2M 4L)
choose one of the 2 fighter cocks(2C1) then frm remaining guys i.e 6 select 1(6C1).Now 2 members hav bn slected .select 4 ladies i.e 5C4

so it becomes 2C1 * 6C1 * 5C4

2nd case:-(2M 4L)
Leave both fighter cocks...choose 2 guys from remaining 6...6C2
4 ladies frm 5 ..i.e 5C4

6C2 * 5C4

3rd case:-(3M 3L)
choose one of the 2 F. C's 2C1 ,then 2 frm remaining 6 6C2 , then 3 ladies from 5 5C3

2C1*6C2*5C3

4th case:-(3M 3L).
no more F.c's..leave them.
6C3 * 5C3

add up all 4....u get 635

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Manager
Joined: 23 Jan 2006
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18 Apr 2006, 11:51
there are 2 possible scenarios 2M,4W and 3M,3W
so,

(2C8 * 4C5) + (3C8 * 3C5) = 140 + 560 = 700 possible combinations
You know that there are invalid cases where the two men are together, so by process of elimination, E (635) is the only valid answer, however I would suspect that the real GMAT would put a few more choices under 700 so you'd have to solve the whole thing.

so 700 total cases, minus the cases where the two jerks are together.
all of comboas that have 2 men that are those two, are invalid, so thats the two of them, paired with every possible set of women, ->5
The 3-men sets that are invalid are the two of them, paired with each of the remaining six men(6), pair each of those with every possible combination of 3 women(10) 6*10 = 60.
so there are 5 invalid combos that have 2 men, and 60 invalid combos that have 3 men.

700 - (5 + 60) = 635

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18 Apr 2006, 11:51
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# A Committee of 6 is chosen from 8 men and 5 women, so as to

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