Ankitaverma wrote:The 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 serve on the committee?
a. 16
b. 24
c. 26
d. 30
e. 32
Another approach is to recognize that:
# of permissible committees =
total # of 3-person committees that ignore the rule -
# of 3-person committees that break the rule
total # of 3-person committees that ignore the rule
If we ignore the rule about married couples, we can select any 3 people from the 8 people.
Since the order of the 3 selected people does not matter, we can use combinations.
We can select 3 people from 8 people in 8C3 ways (=
56 ways)
Aside: If anyone is interested, we have a free video on calculating combinations (like 8C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
# of 3-person committees that break the rule
We want the number of committees consisting of an entire couple and a third person.
Let's take the task of building a 3-person committee and break it into stages.
Stage 1: Select 1 of the 4 couples.
We'll place both people in this couple on the committee.
There are 4 couples, so this stage can be accomplished in
4 ways
Stage 2: Select the third person for the committee
There are now 6 people remaining, so this stage can be accomplished in
6 ways.
By the Fundamental Counting Principle (FCP) we can complete both stages (and thus create a 3-person committee) in
(4)(6) ways
In other words, we can create
24 committees that break the rule.
So, #
of permissible committees =
56 -
24
=
32
=
E
Cheers,
Brent
