a committee of four is formed from five eligible members. let the eligible members be represented with a, b…

a committee of four is formed from five eligible members. let the eligible members be represented with a, b, c, d, and e. the possible outcomes include s = {abcd, bcde, acde, abce, abde}. which statements about the situation are true? check all that apply. there are 120 different ways to choose the committee. if person a must be on the committee, there is only one way to form the committee. if persons a and c must be on the committee, there are three ways to form the committee. there are five ways to form the committee if person e must be on it. if the number of eligible members increases, the number of outcomes increases.

a committee of four is formed from five eligible members. let the eligible members be represented with a, b, c, d, and e. the possible outcomes include s = {abcd, bcde, acde, abce, abde}. which statements about the situation are true? check all that apply. there are 120 different ways to choose the committee. if person a must be on the committee, there is only one way to form the committee. if persons a and c must be on the committee, there are three ways to form the committee. there are five ways to form the committee if person e must be on it. if the number of eligible members increases, the number of outcomes increases.

Answer

Explanation:

Step1: Calculate total number of ways

The formula for combinations is (C(n,r)=\frac{n!}{r!(n - r)!}), where (n = 5) (number of members) and (r=4) (size of committee). So (C(5,4)=\frac{5!}{4!(5 - 4)!}=\frac{5!}{4!1!}=\frac{5\times4!}{4!}=5\neq120). So the first statement is false.

Step2: If (A) must be on the committee

We need to choose (3) more members from (4) ((B,C,D,E)). Using combination formula (C(4,3)=\frac{4!}{3!(4 - 3)!}=\frac{4!}{3!1!}=4\neq1). So the second statement is false.

Step3: If (A) and (C) must be on the committee

We need to choose (2) more members from (3) ((B,D,E)). Using combination formula (C(3,2)=\frac{3!}{2!(3 - 2)!}=\frac{3!}{2!1!}=3). So the third statement is true.

Step4: If (E) must be on the committee

We need to choose (3) more members from (4) ((A,B,C,D)). Using combination formula (C(4,3)=\frac{4!}{3!(4 - 3)!}=\frac{4!}{3!1!}=4\neq5). So the fourth statement is false.

Step5: Effect of increasing eligible members

As the number of eligible members ((n)) increases and (r) (size of committee) is fixed ((r = 4)), since (C(n,r)=\frac{n!}{r!(n - r)!}), when (n) increases, the value of (C(n,r)) increases. So the fifth statement is true.

Answer:

C. If persons A and C must be on the committee, there are three ways to form the committee; E. If the number of eligible members increases, the number of outcomes increases.