. A man has nine friends, four boys and five girls. In how many ways can he invite them, if there have to be exactly
three girls in the invitees?

• 5c3(4c0+4c1+4c2+4c3+4c4)
• 11 years agoHelpfull: Yes(33) No(1)
• 5C3[4C1 +4C2 +4C3 +4C4 ]
• 12 years agoHelpfull: Yes(20) No(29)
• Since we require exactly 3 girls ,therefore total ways are (4C4*5C3)+(4C3*5C3)+(4C2*5C3)+(4C1*5C3)+(4C0*5C3)=160 WAYS
• 8 years agoHelpfull: Yes(2) No(0)
• ans 160
  5c3(1+4C1+4C2+4C3+4C4)
• 8 years agoHelpfull: Yes(1) No(0)
• 9 friends : 5Girls and 4Boys
  Use permutation concept
  Selecting exactly 3 girls out of 5 : 5c3 = 5!/(5-3)!*3! = 10 ways
  Selecting boys : 4C0 + 4C1 + 4C2 + 4C3 + 4C4 = 16 ways
  Total ways = 10*16 = 160
• 8 years agoHelpfull: Yes(1) No(0)
• Out of the 5 girls, 3 girls can be invited in 5C3 ways.
  
  He can invite one, two, three, four or even no boys. Out of 4 boys, he can invite them in the said manner in (2)4 ways.
  
  Thus, the total number of ways is 5C3 × (2)4 = 10 × 16 = 160.
• 7 years agoHelpfull: Yes(1) No(0)
• He can select three girls in 5C3 ways = 10 ways.
  He can select boys in [4C0 + 4C1 + 4C2 + 4C3 + 4C4] ways = 16 ways.
  Total number of ways = 10*16 = 160 Answer.
• 8 years agoHelpfull: Yes(0) No(0)