In how many ways a team of 11 must be selected a team 5 men and 11 women such that the team must comprise of not more than 3 men.

• not more than 3 men be selected:- so, different combinations for the formation of the team will be 3men,8women; 2men,9women; 1men,10 women; 0men, 11 women.
  5C0*11C11+11C10*5C1+11C9*5C2+11C8*5C3
  =1+11*5+55*10+165*10
  =1+55+550+1650
  =2256
• 9 years agoHelpfull: Yes(28) No(0)
• 5c3*11c8+11c11+5c2*11c9+5c1*11c10=2266
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• =>5C0*11C11+11C10*5C1+11C9*5C2+11C8*5C3
  =1+11*5+55*10+165*10
  =1+55+550+1650
  =2256
• 9 years agoHelpfull: Yes(0) No(0)
• 5C3*11C8+5C2*11C9+5C1*11C10+11C11
  Solve this you will get the answer..
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• The question says that there should be atleast three men. So the selection can be done in the following ways-
  11C11+5C1*11C10+5C2*11C9+5C3*11C8
  =2256
• 9 years agoHelpfull: Yes(0) No(0)
• Case 1: Men: 0 and Women: 11 => C(5,0)+C(11,11) = 001
  Case 2: Men: 1 and Women: 10 => C(5,1)+C(11,10) = 055
  Case 3: Men: 2 and Women: 09 => C(5,2)+C(11,09) = 550
  Case 4: Men: 3 and Women: 08 => C(5,3)+C(11,08) =1650
  ------
  Total Number of Ways =2256
  
  As the maximum number of Men to be included in the team is 3 we stop at C(5,3).
• 9 years agoHelpfull: Yes(0) No(0)
• 5C0*11C11+11C10*5C1+11C9*5C2+11C8*5C3
  =1+11*5+55*10+165*10
  =1+55+550+1650
  =2256
• 9 years agoHelpfull: Yes(0) No(0)
• we can select (0 men & 11 women ) or (1 men & 10 women) or (2 men & 9 women) or (3 men and 8 women)
  number of ways of selection made=1+55+550+1650=2256
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• M W
  1 10
  2 9
  3 8
  SO, 5C1*11C10+5C2*11C9+5C3*11C8=2250
• 9 years agoHelpfull: Yes(0) No(3)