Combination

[ketkoag]

If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
A. 20
B. 40
C. 50
D. 80
E. 120

OA : D
How??

[relic]

Well, 5 couples means 10 people. So need a group of 3 from 10, which becomes 10!/(3!*7!) = 120 total possible groups.

We are not interested in having partners in the group however, so we must remove these occurrences. Each couple could be paired with 8 other people, so for each of the 5 couples there are 8 unacceptable groups, or 40 unacceptable groups.

120 - 40 = 80

So there are 80 acceptable groups.

[ketkoag]

Please explain me why 10!/(3!*7!) ??

[lilu]

B/c you need to choose 3 people out of 10

[vittalgmat]

Here is a simpler way to solve the "married couples" class of problems. Thanks to Ron Purewal.

5 married couples => 10 ppl.

The first slot can be filled in 10 ways.
Second slot can be filled in 8 ways (ie. exclude the person in slot 1 and his/her spouse)

third slot can be filled in 6 ways (exclude the person and their spouses in slots 1 and 2)

Total arrangements = 10*8*6

Now we are looking for combination of 3 ppl. The above reflects permutations of grp of 3 ppl. To get the combination, we divide by 3!
ie
10*8*6/3! = 80

Ht helps