Answer:
In 10,000 ways at least one of the junior partners can be chosen to be on the committee
Step-by-step explanation:
Combinations are used when three people ordered different are the same outcome. For example, choosing Ed, Tremaine and Tre'Davious is the same as choosing Tremaine, Tre'Davious and Ed.
[tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In how many ways can at least one of the junior partners be chosen to be on the committee?
A firm has 16 senior and 25 junior partners.
The total number of committees is:
[tex]C_{41,3} = \frac{41!}{3!(38)!} = 10660[/tex]
The number of committees with no junior partners is:
[tex]C_{16,3} = \frac{16!}{3!(13)!} = 560[/tex]
So the number of committees with at least one junior partner is:
10660 - 560 = 10000
In 10,000 ways at least one of the junior partners can be chosen to be on the committee
At least one of the junior partners could be chosen to be on the committee in 9600 different ways.
Since a firm has 16 senior and 25 junior partners, and a committee of three partners is selected at random to represent the firm at a conference, to determine in how many ways can at least one of the junior partners be chosen to be on the committee, the following calculation must be made:
Therefore, at least one of the junior partners could be chosen to be on the committee in 9600 different ways.
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