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

Answer 1

Answers are 
there are four ways to choose the committee
there are three ways to form the committee if person D must be on it
if person B and C must be on the committee, there are two ways to form the committee

Answer 2

Answer:

The correct statement are:

• There are four ways to choose the committee.
• There are three ways to form the committee if person D must be on it.
• If persons B and C must be on the committee, there are two ways to form the committee.

Step-by-step explanation:

It is given that:

A membership committee of three is formed from four eligible members.

1)

There are four ways to choose the committee.

This statement is true.

since we have to choose 3 members out of the 4 members so we can use the method of combination.

i.e. $\binom{4}{3}=\dfrac{4!}{3!\times (4-3)!}\\\\ \binom{4}{3}=4$

2)

There are three ways to form the committee if person D must be on it.

This statement is also true.

since D has to be in the committee, this means we have to choose 2 more people out of the three people to form the committee.

i.e. $\binom{3}{2}=\dfrac{3!}{2!\times (3-2)!}\\\\\binom{3}{2}=3$

3)

If seven members are eligible next year, then there will be fewer combinations.

This statement is wrong.

Since we have to choose 3 members out of 7 members so the number of possible combinations will be:

$\binom{7}{3}=\dfrac{7!}{3!\times (7-3)!}\\\\\binom{3}{2}=\dfrac{7!}{3!\times 4!}=35$

i.e. there are 35 combinations possible.

4)

If persons B and C must be on the committee, there are two ways to form the committee.

if B and C have to be in the committee then we have to choose just one person out of the two people left.

So possible combination will be:

$\binom{2}{1}=\dfrac{2!}{1!\times (2-1)!}\\\\\binom{2}{1}=2$

Hence, the statement is true.

5)

If persons A and C must be on the committee, then there is only one way to form the committee.

If A and C have to be in the committee then as in last option we have to choose any one of the two person left.

so possible number of ways are 2.

Hence, the statement is false.