GMATPrep1 - DS - Combinations

[hja379]

A committee of 3 people is to be chosen from 4 married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee.

a) 16
b) 24
c) 26
d) 30
e) 32

OA e

[anshumishra]

A committee of 3 people is to be chosen from 4 married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee.

a) 16
b) 24
c) 26
d) 30
e) 32

OA e

Method 1 :

No. of ways to select 3 couples out of 4, who will send their representative = 4C3=4.

Each of these 3 couples can send two persons(wife or husband), so no. of ways = 2*2*2=8.

Total no. of ways of ways: 4C3*8=32. E

Method 2:

First member can be chosen in 8 ways
Second member can be chosen in 6 ways
Third member can be chosen in 4 ways

So, total no. of ways in which the committee can be selected = 8*6*4 ways

This is like arranging the 3 places, but in our case, the order doesn't matters (since if the same set of 3 people who appear in the committee on different places constitute a single committee), so lets's get rid out of the repetitions by diving by 3!

So, the required no. of ways = 8*6*4/3! = 32.


Method 3 :

(M1,F1),(M2,F2),(M3,F3),(M4,F4)

3 People have to be selected (exclude married couple)

All 3 male : 4C3 = 4
All 3 female : 4C3 = 4
2 Male, 1 female : 4C2*2C1 = 6*2 = 12
2 Female, 1 male : 4C2*2C1 = 6*2 = 12

So, total no. of ways = 4+4+12+12 = 32.