asked on 13/09/2015
There are 21 students in a class where boys number exceeds girls no by 1 .
A quiz group of 4 students has to be organized where all 4 should not be from same gender . Then how many ways quiz groups can be created
Options I dnt remember but option B was like 5445

• b+g=21
  b-g=1
  ----------
  2g=20
  g=10,
  b=11
  So,Total Combination are
  bbbg,bbgg,bggg
  (11c3 * 10c1+11c2*10c2+11c1*10c3)
  1650+2475+1320=5445
  
• 9 years agoHelpfull: Yes(53) No(0)
• Ans = Total no of ways - No of ways of groups of same gender
  
  = 21C 4 - { 11C 4 + 10C 4 }
  
  = 5445
• 9 years agoHelpfull: Yes(23) No(1)
• b) 5445
  x= no of boys, y= no of girls
  x+y=21....................................(1)
  x-y=1 ....................................(2)
  Solve these x=11,y=10
  Now Group of 4 where all 4 are not from the same gender can be..
  1) 1Boy3Girls= 11c1*10c3
  2) 2Boys2Girls= 11c2*10c2
  3) 3Boys1Girl= 11c3*10c1
  Hence total ways will be = 11c1*10c3+11c2*10c2+11c3*10c1=1320+2475+1650=5445
  So, If You have guessed b) It's Right.. :)
  
• 9 years agoHelpfull: Yes(6) No(0)
• total girls are 10 and boys are 11 so no of ways are 10c1*11c3+10c2*11c2+10c3*11c1
  which is 1650+2475+1320=5445 &b IS THE ANSWER
• 9 years agoHelpfull: Yes(1) No(0)
• as per question, 11 boys and 10 girls ...
  Total number of ways of groups created by 21 peolpes = 21c4=5985
  # of ways in creating grp consiting of only men = 11c4=330
  # of ways in creating grp consiting of only women = 10c4=210
  The reqd # of ways = 5985-330-210=5445
• 9 years agoHelpfull: Yes(1) No(0)
• ans-5445
  soln-
  boy=11
  girl=10
  
  no of ways =[3boy1girl]+[2boy2girl]+[1boy3girl]
  = [11C3*10C1+11C2*10C2]+[11C1*10C3]
  =5445
• 9 years agoHelpfull: Yes(0) No(0)
• 11 boys and 10 girls ...
  11c3 * 10c1 + 11c2*10c2 + 11c1*10c3 = 8470
• 9 years agoHelpfull: Yes(0) No(2)
• Answer is probably option b 5445.
  21= x + (x-1), where x is no. of boys.
  Hence 11 boys and 10 girls.
  Now, (11C3 * 10C1) + (11C2*10C2) + (11C1 * 10C3) =5445.
• 9 years agoHelpfull: Yes(0) No(0)
• 5445
  given no.of boys is 1 more than girl .
  let no. of boys be n
  then no of girl is n-1
  a/q
  n+n-1=21
  n=11
  now three cases arises
  case 1: selecting 2 boys and 2 girls
  11c2*10c2
  case 2: selecting 3 boys and 1 girls
  11c3*10c1
  case 3: selecting 1 boys and 3 girls
  11c1*10c3
  as the above cases is mutually exclusive so
  total no of selection i=11c2*10c2+11c3*10c1+11c1*10c3=5445
  
  
  
• 9 years agoHelpfull: Yes(0) No(0)
• Hi Chetan.. You got your result for 13/09/2015 exam?
• 9 years agoHelpfull: Yes(0) No(1)
• Boys=11, girls =10;
  21 students can be arranged in 21c4 ways = 5985 ways
  11 boys can ...... in 11c4ways = 330 ways
  10 girls can ...... in 10c4=210 ways
  There a group without all members of same gender = (total ways)-(only boys)-(only girls)
  =5985-330-210=5445
• 9 years agoHelpfull: Yes(0) No(0)
• Proper explanation From this concept you can solve any case
  
  Total 6 numbers are there 1,2,3,4,5,6 among them we have to make 5 digits
  04= rest remaining 4 so they have to fit in 3 slots!! so ways is 4!
  20= rest remaining 4 so they have to fit in 3 slots!! so ways is 4!
  40= rest remaining 4 so they have to fit in 3 slots!! so ways is 4!
  12= rest remaining 4 so they have to fit in 3 slots!! but if 0 comes in first place it will not make 5 digit numbers...so we have to choose from 3 so ways are 3!*3=18
  32=12= rest remaining 4 so they have to fit in 3 slots!! but if 0 comes in first place it will not make 5 digit numbers...so we have to choose from 3 so ways are 3!*3=18
  24=12= rest remaining 4 so they have to fit in 3 slots!! but if 0 comes in first place it will not make 5 digit numbers...so we have to choose from 3 so ways are 3!*3=18
  52=12= rest remaining 4 so they have to fit in 3 slots!! but if 0 comes in first place it will not make 5 digit numbers...so we have to choose from 3 so ways are 3!*3=18
  therefore add all cases=4!+4!+4!+18+18+18+18=144 ans.
  
• 8 years agoHelpfull: Yes(0) No(2)