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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gmatnmein2010
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harsh.champ
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Taking casesgmatnmein2010 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
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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thephoenix
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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
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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Warlock007
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abe Case 1 mein 8C1 ki jagah 6C1 kar mere bhaiharsh.champ wrote:Taking casesgmatnmein2010 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
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
baaki sab theek hai
jiyoooooooo
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gmatter2012
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Not able to get the logic behind , selecting two men who work together with each other.
can anybody explain , please?
can anybody explain , please?
