in how man was team of four can be formed from four boys and three girls such that atleast one boy and one girl should be there?

• 34 ways
  
  One boy and one girl can be selected as (one boy & three girls) or (two boys & two girls) or (three boys & one girl)=(4C1*3C3)+(4C2*3C2)+(4C3*3C1)=4+18+12=34
• 8 years agoHelpfull: Yes(42) No(1)
• 4c1*3c3+4c2*3c2+4c3*3c1=34
• 8 years agoHelpfull: Yes(5) No(0)
• The possible cases could be
  1. 3B and 1G = 4*3=12
  2. 2B and 2G = 6*3 = 18
  3. 1B and 3G = 4*1 =4
  
  ANSWER =12+18+4 =34
• 8 years agoHelpfull: Yes(5) No(0)
• 7C4-4C4
  
  Total No of ways is 7C4 in that
  no of ways of having all boys is 4C4
  and in all remaining ways we have atleast one boy and one girl
  
• 8 years agoHelpfull: Yes(3) No(3)
• (4C1 * 3C3) + (4C2 * 3C2) + (4C3 + 3C1)
• 7 years agoHelpfull: Yes(2) No(0)
• ans is 34/35
• 8 years agoHelpfull: Yes(1) No(2)
• since 1B and 1G has to be there, so remaining boys =3 and remaining girls=2.
  now, no. of ways of selecting on 2 seats out of 3 boys and 2 girls is (3c1*2c1) +(3c2*2c0) +(3c0*2c2).
• 5 years agoHelpfull: Yes(1) No(0)
• 4C1*3C1*5C2=120
• 8 years agoHelpfull: Yes(0) No(9)
• 44/105
  total number of ways :- 7C4
  since atleast 1boy and 1 girl has to be selected : -( 4C1*3C3+4C2*3C2+4C3*3C1)/7C4
• 8 years agoHelpfull: Yes(0) No(6)
• Answer will be =144
  First we allow women to sit by 3!
  and then by 4! men will be inserted in between them and corners.
  So finally 4!*3!=144
• 7 years agoHelpfull: Yes(0) No(3)
• ncr=7c4=35:
  r=group of 4,n=total no of man
  35-4c1*3c1=35-1=34
• 5 years agoHelpfull: Yes(0) No(0)
• 16
  4c1*3c1+4c2+3c2
• 5 years agoHelpfull: Yes(0) No(0)