A committee of six is selected from ten people, of whom A and B are two. How many committees can
be formed:
(a) containing both A and B
(b) excluding A if B is included?
To solve this problem, we need to use the concept of combinations.
(a) Containing both A and B:
Since both A and B must be included in the committee, we have already selected 2 people. We need to select the remaining 4 committee members from the remaining 8 people.
The number of ways to select 4 people from a group of 8 can be found using the combination formula:
C(8, 4) = 8! / (4! * (8-4)!) = 70
So, there are 70 committees that can be formed containing both A and B.
(b) Excluding A if B is included:
If B is included, we have already selected 1 person. We need to select the remaining 5 committee members from the remaining 9 people (excluding A and B).
The number of ways to select 5 people from a group of 9 can be found using the combination formula:
C(9, 5) = 9! / (5! * (9-5)!) = 126
So, there are 126 committees that can be formed excluding A if B is included.