there are 41 students in a class, number of girls is one more than number of guys. we need to form a team of four students. all four in the team cannot be from same gender. number of girls and guys in the team should NOT be equal. How many ways can such a team be made?

• Your approach is right mallikarjuna, but 21C3*20C1 + 20C3*21C1 is not 49400. its 50540. :)
• 8 years agoHelpfull: Yes(31) No(0)
• boys=20 and girls=21 now the combinations are {girl,girl,girl,boy} or {boy,boy,boy,girl}
  so 21c3*20c1+20c3*21c1=49,400 ways.
• 8 years agoHelpfull: Yes(22) No(11)
• B=20
  G=21
  As number of them cannot be same and cannot be filled by any one group
  So total number of ways>>
  ( 20 C 3 * 21 C 1) + ( 20 C 1 * 21 C 3)= 50540
• 8 years agoHelpfull: Yes(5) No(0)
• G + B= 41 and B=G-1 hence, G=21 and B=20
  no. of ways = 21C1*20C3 + 20C1*21C3 = 50540
• 8 years agoHelpfull: Yes(2) No(0)
• If no of girls is 1 more than boys then no of boys be X and no of girls be X+1. According to question total is 41 , so (X+1)+x=41.
  2X=40, then X=20.
  Means Boy=20 and girl=21.
  Now according to question only 2 case is possible (i.e., 1boy+3girl or 1girl+3boy).
  such team formed in
  case1-: 20c1*21c3
  case2-: 21c1*20c3
  By solving case1 and case2, I have (20*7*10*19) + (21*10*19*6) = 50540
• 7 years agoHelpfull: Yes(2) No(0)
• 21C1*20C3 + 20C1*21C3 = 50540
• 8 years agoHelpfull: Yes(1) No(0)
• 41C4-(21c4+20c4+21c2*20c2)=91040
• 7 years agoHelpfull: Yes(1) No(0)
• There can be more cases na?no of boys and girls can be 22,19;23.18.....??
• 8 years agoHelpfull: Yes(0) No(8)
• To form team we need to perform combination(i.e. selection),
  For , G to be more than 1 compared to M , it should be G=21 and M=20,
  Now we have 2 options,
  1G and 3M
  or
  3G and 1M
  (2G and 2M or 0G and 4M or 4G and oM are not allowed),
  
  Total : 21c1 * 20c3 + 21c3*20c1 = 50540 ways.
  
• 8 years agoHelpfull: Yes(0) No(1)