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?

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 = [tex]{51 \choose 3} = \frac{51!}{3!(51-3)!} = 20,825[/tex]

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 = [tex]{25 \choose 3} = \frac{25!}{3!(25-3)!} = 2,300[/tex]

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

ayokenny2001 ayokenny2001 · Answer 2 · 2022-03-24T12:55:40+01:00

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

What is combination in statistics?

A combination is a selection of all or part of a set of objects, without regard to the order in which objects are selected.

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

= ( 5/3 )

= 5! / 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 / 3)

= 25! / 3! ( 23 - 3) !

= 2,300

Hence, 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

Learn more about combinations here : https://brainly.com/question/4658834