Carcass wrote:
A committee of three people is to be chosen from four 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
Let's Suppose we have 4 Couples.
Call them
M1 F1
M2 F2
M3 F3
M4 F4
Now, as per question statement, apply FCP:
we have three Slots
____ ____ ____
How many choices do we have for Slot 1? 8 Choices because any of the person from a couple can be selected for 1st Slot
=>
8Now move on to the 2nd Slot. How many choices do we have for the 2nd Slot? Let's say we have selected M1 for Slot 1. Now, F1 can't be selected for 2nd Slot because of the restriction in the question statement.
For 2nd Slot, total number of choices will be
8 6Similary, for 3rd Slot, Let's say we have selected F2 for Slot 2. Now, M2 can't be selected for 3rd Slot because of the restriction in the question statement.
Since, M1 and F2 are already selected by us for slot 1 and 2 respectively, F1 and M2 can't be selected for third slot, and M1 and F2 are already on the committee. We are left with only four choices which are: M3 F3 M4 F4
So, for Slot 3
8 6 4Now, divide the whole expression by 3! to remove the repeatitions.
=> 8*6*4/6
=> 32
Hence E!