Carcass wrote:
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
Using a more indirect method, you can subtract out the undesirable outcomes from the total number of outcomes to get the desired outcomes.
In this case, the undesired outcome is having a married couple out of the three spots in the committee.
There are a total of 10 people to choose from (5 couples).
Order doesn't matter (A group of Person 1, Person 2, and Person 3 is the same group as Person 2, Person 1, and Person 3), so we use combinations.
10 Choose 3 = \(\frac{10*9*8}{3*2} = 5*3*8 = 120\)
So if there were no restrictions, the total number of outcomes would be 120 different committees.
Now consider the undesired outcomes: choosing a married couple for 2 of the 3 spots on the committee.
Since there are 5 couples, if we choose 1 couple (2 people) to fill two spots, we could choose any of the remaining 8 people to fill the third spot.
That's 8 different combinations for each of the 5 couples. So there are 40 different committees that include a married couple.
The total number of undesired outcomes then is 40.
Now we subtract: 120-40 = 80,
giving us our answer of D.