Question

A membership committee of three is formed from four eligible members. Let the eligible members be represented by A, B, C, and D. The possible outcomes include S = {ABC, ABD, ACD, BCD}.

Which statements about the situation are true? Check all that apply.

A. There are four ways to choose the committee.
B There are three ways to form the committee if person D must be on it.
C If seven members are eligible next year, then there will be fewer combinations.
D If persons B and C must be on the committee, there are two ways to form the committee.
E If persons A and C must be on the committee, then there is only one way to form the committee.

which ones do i check??????

Answer 1

Options A, B and D are correct.

Explanation:

A membership committee of three is formed from four eligible members. Let the eligible members be represented by A, B, C, and D. The possible outcomes include S = {ABC, ABD, ACD, BCD}.

The true statements about this situation are :

A. There are four ways to choose the committee. As we can see 'S' has 4 ways.

B. There are three ways to form the committee if person D must be on it. We have three sets that have D.

D. If person B and C must be on the committee, there are two ways to form the committee. Yes, B and C are only in 2 sets.

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C. If seven members are eligible next year, then there will be fewer combinations. This is wrong as there will be more combinations like 7C6 if 6 are to be chosen out of 7.

E If persons A and C must be on the committee, then there is only one way to form the committee. This is wrong as we have two sets that have A and C in it.

Answer 2

Choosing 3 persons out of the four persons available in for the committee, we use the concept of combination,
4C3 = 4
The statements that are true for this item are: A, B, and D.