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by gmatnmein2010 » Mon Feb 08, 2010 5:36 am
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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by harsh.champ » Mon Feb 08, 2010 5:58 am
gmatnmein2010 wrote: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
Taking cases
Case 1:- 3men,3 women 8C3x 5C3 - 8C1 x 5C3
Case 2 :-2 men,4 women 8C2 x 5C4 - 6C0 x 5C4

Remove the cases when those 2 men are there
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by thephoenix » Mon Feb 08, 2010 6:45 am
Two cases: 2 men and 4 women OR 3 men and 3 women.
Ways to chose 6 members committee without restriction (two men refuse to server together)=8C2*5C4+8C3*5C3 = 700
Ways to chose 6 members with two men serve together=2C2*5C4+2C2*6C1*5C3=5+60=65
700-65 = 635

imo e
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by Warlock007 » Thu Mar 10, 2011 12:07 pm
harsh.champ wrote:
gmatnmein2010 wrote: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
Taking cases
Case 1:- 3men,3 women 8C3x 5C3 - 8C1 x 5C3
Case 2 :-2 men,4 women 8C2 x 5C4 - 6C0 x 5C4

Remove the cases when those 2 men are there
abe Case 1 mein 8C1 ki jagah 6C1 kar mere bhai
baaki sab theek hai
jiyoooooooo
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by gmatter2012 » Sat Sep 15, 2012 6:08 am
Not able to get the logic behind , selecting two men who work together with each other.
can anybody explain , please?
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