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? select three options. 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.

Explanation:

Step1: Analyze the number of ways to choose the committee

Given \(S=\{ABC, ABD, ACD, BCD\}\), the number of elements in \(S\) is \(4\). So, there are \(4\) ways to choose the committee.

Step2: Analyze the case when \(D\) must be on the committee

If \(D\) is fixed, we need to choose \(2\) more members from \(\{A,B,C\}\). The combinations are \(\{ABD,ACD,BCD\}\), so there are \(3\) ways.

Step3: Analyze the case when \(B\) and \(C\) must be on the committee

If \(B\) and \(C\) are fixed, we need to choose \(1\) more member from \(\{A,D\}\). The combinations are \(\{ABC,BCD\}\), so there are \(2\) ways.

Step4: Analyze the case when \(A\) and \(C\) must be on the committee

If \(A\) and \(C\) are fixed, we need to choose \(1\) more member from \(\{B,D\}\). The combinations are \(\{ABC,ACD\}\), so there are \(2\) ways (not \(1\) way).

Step5: Analyze the case when there are \(7\) eligible members

The formula for combinations is \(C(n,k)=\frac{n!}{k!(n - k)!}\). For \(n = 4,k=3,C(4,3)=\frac{4!}{3!(4 - 3)!}=4\). For \(n = 7,k = 3,C(7,3)=\frac{7!}{3!(7-3)!}=\frac{7\times6\times5}{3\times2\times1}=35\). So, there will be more combinations when \(n = 7\).

Answer:

• 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.