Question

A firm has 25 senior and 26 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:

There are 18,525 ways to choose at least one of the junior partners to be on the committee.

Explanation:

If all partners can be chosen in any way the choices would be combinations of any 3 partners from all 51 partners. The number of ways to arrange it would be:

A = ${51 \choose 3} = \frac{51!}{3!(51-3)!} = 20,825$

The opposite of choosing at least one junior partner is to choose no junior partner. Those choices would be combinations of any 3 partners from 25 senior partners. The number of ways to arrange it would be

B =  ${25 \choose 3} = \frac{25!}{3!(25-3)!} = 2,300$

So the number of ways to choose at least one of the junior partners to be on the committee would be: A - B = 20,825 - 2,300 = 18,525