Question

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?

Answer 1

Answer:

A: $\binom{8}{4}$

B:  $\binom{9}{6} + \binom{8}{5}$

Step-by-step explanation:

Part A

Lets describe the process of forming a committee with both A and B:

• First step, add A and B to the committee, there is 1 way to do this.
• Second step, choose the remaining 4 committee members from the remaining 8 people. There is $\binom{8}{4}$ (8 choose 4) ways to do this.

Now we use the product rule and say that the number of ways to create a committee as described in part A is: $1 \times \binom{8}{4} = \binom{8}{4}$

Part B

Lets describe the process of forming a committee excluding A if B is in the committee.

First lets do when B is not in the committee. (So A is allowed in the committee): This is one step, we choose 6 members from 9 potential members, there is $\binom{9}{6}$ ways to do this.

Second situation is when B is in the committee. (So A is NOT allowed in the committee:

• Place B in the committee, there is 1 way to do this.
• Choose the remaining 5 spots, from 8 possible people. $\binom{8}{5}$ ways to do this.

Now since the first and second situation are mutually exclusive (they have no overlap), we can simply sum the ways of doing it in each situation. So the final answer for part B is: $\binom{9}{6} + \binom{8}{5}$