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?

• possible combinations can be either 2M+4W or 3M+3W
  
  Total possible combinations = 8c2*5c4 + 8c3*5c3 = 700
  Combinations in which both men are together = 5c4 + 6c1*5c2 = 65
  
  Combination in which they are not together = 700-65 = 635
• 11 years agoHelpfull: Yes(25) No(2)
• wo 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
• 11 years agoHelpfull: Yes(6) No(2)
• totel no of such committee =(m=2,w=4) and (m=3,w=3) only two committe form .
  no of different committees=( (8c2-6c0)*5c4 + (8c3-6c1)*5c3 )= 635
  sorry , last time my calculation is wrong but method is right .
• 11 years agoHelpfull: Yes(4) No(2)
• totel no of such committee =(m=2,w=4) and (m=3,w=3) only two committe form .
  no of different committees=( (8c2-6c0)*5c4 + (8c3-6c1)*5c3 )= 830
• 11 years agoHelpfull: Yes(2) No(7)
• correct ans in 500 +135=635
• 11 years agoHelpfull: Yes(2) No(7)
• 635 is the right ans......
• 11 years agoHelpfull: Yes(1) No(1)
• 8c26c4+8c35c3-7=913 ans
  
• 11 years agoHelpfull: Yes(0) No(17)