Question

A firm has 18 senior and 22 junior partners. A committee of three partners is selected at random to represent the firm at a conference. In how many ways can at least one of the junior partners be chosen to be on the committee?

Answer 1

Answer:

Answer is 24288.

Step-by-step explanation:

Given that there are 18 senior and 22 junior partners.

To find:

Number of ways of selecting at least one junior partner to form a committee of 3 partners.

Solution:

At least junior 1 member means 3 case:

1. Exactly 1 junior member

2. Exactly 2 junior member

3. Exactly 3 junior member

Let us find number of ways for each case and then add them.

Case 1:

Exactly 1 junior member:

Number of ways to select 1 junior member out of 22: 22

Number of ways to select 2 senior members out of 18: 18 $\times$ 17

Total number of ways to select exactly 1 junior member in 3 member committee: 22 $\times$ 18 $\times$ 17 = 6732

Case 2:

Exactly 2 junior member:

Number of ways to select 2 junior members out of 22: 22 $\times$ 21

Number of ways to select 1 senior member out of 18: 18

Total number of ways to select exactly 2 junior members in 3 member committee: 22 $\times$ 21 $\times$ 18 = 8316

Case 3:

Exactly 3 junior member:

Number of ways to select 3 junior members out of 22: 22 $\times$ 21 $\times$ 20 = 9240

So, Total number of ways = 24288