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by Anju@Gurome » Tue Mar 19, 2013 3:21 am
vipulgoyal wrote:IN HOW MANY WAYS CAN U SELECT A COMMETTEE OF 4 PEOPLE FROM A GROUP OF 5 COUPLES SO THAT NO COUPLE IS CHOSEN?
Each couple can send only one "representative" to the committee.
Now, number of ways to select 4 couples (as there should be 4 members) who will send only one "representative" to the committee out of 5 couples = 5C4 = 5

Each of these 4 chosen couples can send two persons, either the husband or the wife
Hence, for each of the above combinations, number of possible combinations are = 2*2*2*2 = 16

Hence, number of ways to choose 4 people out 5 couples so that no couple is chosen = 5*16 = 80
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by vipulgoyal » Tue Mar 19, 2013 4:20 am
Thanks, very helpful post as always
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by GMATGuruNY » Tue Mar 19, 2013 9:50 am
vipulgoyal wrote:IN HOW MANY WAYS CAN U SELECT A COMMETTEE OF 4 PEOPLE FROM A GROUP OF 5 COUPLES SO THAT NO COUPLE IS CHOSEN?
An alternate approach:
Number of options for the first person = 10. (Any of the 10 people.)
Number of options for the second person = 8. (Of the 9 remaining people, anyone but the mate of 1 person already chosen.)
Number of options for the third person = 6. (Of the 8 remaining people, anyone but the mates of 2 people already chosen.)
Number of options for the fourth person = 4. (Of the 7 remaining people, anyone but the mates of the 3 people already chosen.)
To combine these options, we multiply:
10*8*6*4.
Since the order of the selections doesn't matter -- ABCD is the same committee as BDAC -- we divide by the number of ways to ARRANGE the 4 people (4!):
(10*8*6*4)/(4*3*2*1) = 80.
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