Latest elitmus Aptitude Question SOLUTION: 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....

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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....

Asked In elitmus ranjit mishra (11 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

Sharathkumar Anbu  (11 years ago)

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.

Is this solution Helpfull? Yes (45) | No (1)  

mahima srivastava  (11 years ago)

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

Is this solution Helpfull? Yes (13) | No (8)  

Tarun Sharma  (11 years ago)

combination of no couple = total-when one couple together
=10c3-5c1*8c1
=120-40
=80.

Is this solution Helpfull? Yes (11) | No (2)  

jagadeesh kumar.B  (11 years ago)

(5c3+5c3)*(5c1+4c2)*(5c2+3c1) = 80

Is this solution Helpfull? Yes (3) | No   

N Arun Singh  (11 years ago)

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

Is this solution Helpfull? Yes (1) | No (1)  

swethakottakota  (11 years ago)

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

Is this solution Helpfull? Yes (1) | No   

bhanu prakash reddy a  (11 years ago)

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

Is this solution Helpfull? Yes (1) | No (5)  

Gaurav Malhotra  (10 years ago)

its simple 5C3 + 5C3 + 5C1*4C2 + 5C2*3C1

Is this solution Helpfull? Yes (1) | No   

moumita ghose  (10 years ago)

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

Is this solution Helpfull? Yes (1) | No   

harshit chowdary  (11 years ago)

Actually the options are 60,80,115,120

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nagireddy  (11 years ago)

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

Is this solution Helpfull? Yes | No (2)  

nandish  (11 years ago)

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

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basanagouda  (10 years ago)

@ Tarun sharma :
ur right that is the shortest method to get answer as 80.

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veer kumar  (10 years ago)

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

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Himanshu Srivastava  (10 years ago)

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

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