Q. There are 5 couple out of which we have to form a 3 member committee so that no couple will be together

Option
a)110
b)120
don't know the other two....

• There are totally 10 members( 5 couples). No couple should come together. So,
  No. of ways to select 1st member is 10.
  No. of ways to select 2nd member is 8. (8 because we should not include the spouse or husband of the 1st member)
  No. of ways to select 3rd member is 6. Same reason stated as above.
  We have to find the number of combinations possible.
  n C r = n P r / r !
  For our problem n P r is 10*8*6 ;
  r = 3;
  Solution : 10*8*6/3!
  So 80 ways.
• 10 years agoHelpfull: Yes(45) No(1)
• ways of selecting three couples= 5c3
  ways of selecting a person from a couple=2
  ways of selecting a person from three couples= 2*2*2
  ways of selecting 3 members for committee= 5c3*2*2*2=80
  so total 80 ways
• 10 years agoHelpfull: Yes(13) No(8)
• combination of no couple = total-when one couple together
  =10c3-5c1*8c1
  =120-40
  =80.
• 10 years agoHelpfull: Yes(11) No(2)
• (5c3+5c3)*(5c1+4c2)*(5c2+3c1) = 80
• 10 years agoHelpfull: Yes(3) No(0)
• question is no couple is selected in team ...
  total selections =10c3 =120
  selecting 1 couppe is 5c1 =5 ...
  total no couple is seleted is 120-5=115 ...options
  were 60,80,115,120 ...can any one verify my ans
• 10 years agoHelpfull: Yes(1) No(1)
• out of 10, we should select 3, so 10c3
  if wethere should be 1 couple, so 5c1, after taking 1 couple out, we have 8 persons are left, so probability of having 1 couple in the 3 member committee is,,
  5c1 * 8c1 /10c3 = 40/120 = 1/3
  probability that having no couple in member committee = 2/3
• 10 years agoHelpfull: Yes(1) No(0)
• total 10 members, we need to form 3 members committee.
  we will select first member in 10c1 ways
  after selecting first person then we need to select 2nd member out of 8 because we should not select spouse of first member.this can be done in 8c1 ways.
  similarly 3rd member can be selected in 6c1 ways.
  
  total=10*8*6=480
  
• 10 years agoHelpfull: Yes(1) No(5)
• its simple 5C3 + 5C3 + 5C1*4C2 + 5C2*3C1
• 10 years agoHelpfull: Yes(1) No(0)
• total 5 couples dey can be considered as 5 grups..each having 2 members. now frm 5 grup 3 pepl cn be selected in 5C3 ways..an each group has 2 members so 2 ways of selecting from individual grup. so 5C3X2X2X2=80
  
• 10 years agoHelpfull: Yes(1) No(0)
• Actually the options are 60,80,115,120
• 10 years agoHelpfull: Yes(0) No(0)
• 5 couple=10 people
  so we have to select 3 member committe
  10c3=120
  but it should be from not couplee so 120-5=115
  115 is the ans
  i think this is my choice not exactly correct
  
• 10 years agoHelpfull: Yes(0) No(2)
• answer .B.120
  Suppose AB,CD,EF,GH,IJ are couples,
  then male side A,C,E,G,I and female side B,D,F,H,J..
  'A' can make with C,D,F,H taken two at a time or he can make with the male side itself i.e.C,E,G,I taken two at a time.so he makes 12.totaly there are 10 members.
  so answer 10*12=120.
  
  so A can go with ways
• 10 years agoHelpfull: Yes(0) No(0)
• @ Tarun sharma :
  ur right that is the shortest method to get answer as 80.
  
• 10 years agoHelpfull: Yes(0) No(0)
• x=2^30=2^(3*10)=8^10 in the same way y=9^10 and z=6^10 from here we came to know which is greater and which is least and next its very easy to predict by removing the brackets in option and all have xyz
• 10 years agoHelpfull: Yes(0) No(0)
• 5 couples
  if comitee of 3 is to be created
  then possibilities=any 3 men or any 3 women or any 2 men and 1 woman or 2 women 1man
  5c3+5c3+5c2*5c1+5c1*5c2
  =2*5c3+2*5c2*5c1=120
• 10 years agoHelpfull: Yes(0) No(0)