Question

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.

Explanation:

Step1: Analyze the total number of ways to choose the committee

The problem is about combinations (since the order of selection doesn't matter for a committee). The formula for combinations is \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 5 \) (eligible members) and \( k = 4 \) (committee size).

\[ C(5,4)=\frac{5!}{4!(5 - 4)!}=\frac{5!}{4!1!}=\frac{5\times4!}{4!×1}= 5 \]
So the first statement "There are 120 different ways to choose the committee" is false because the actual number is 5, not 120.

Step2: Analyze the case when person A must be on the committee

If A must be on the committee, we need to choose 3 more members from the remaining 4 (B, C, D, E). Using the combination formula with \( n = 4 \) and \( k = 3 \):

\[ C(4,3)=\frac{4!}{3!(4 - 3)!}=\frac{4!}{3!1!}=\frac{4\times3!}{3!×1}=4 \]
So the statement "If person A must be on the committee, there is only one way to form the committee" is false.

Step3: Analyze the case when persons A and C must be on the committee

If A and C must be on the committee, we need to choose 2 more members from the remaining 3 (B, D, E). Using the combination formula with \( n = 3 \) and \( k = 2 \):

\[ C(3,2)=\frac{3!}{2!(3 - 2)!}=\frac{3!}{2!1!}=\frac{3\times2!}{2!×1}=3 \]
So the statement "If persons A and C must be on the committee, there are three ways to form the committee" is true.

Step4: Analyze the case when person E must be on the committee

If E must be on the committee, we need to choose 3 more members from the remaining 4 (A, B, C, D). Using the combination formula with \( n = 4 \) and \( k = 3 \):

\[ C(4,3)=\frac{4!}{3!(4 - 3)!}=4 \]
Wait, but looking at the sample space \( S=\{ABCD, BCDE, ACDE, ABCE, ABDE\} \), when E is in the committee, the subsets are BCDE, ACDE, ABCE, ABDE. Wait, no, wait: if E must be in the committee, we choose 3 from the other 4. But the sample space given has 5 elements, each with 4 members, and when E is in the committee, let's list them: BCDE (E is in), ACDE (E is in), ABCE (E is in), ABDE (E is in), and what about ACD E? Wait, no, the sample space is all 4 - member subsets of 5 elements. The number of 4 - member subsets containing E is equal to the number of 3 - member subsets of the remaining 4 elements (A, B, C, D). \( C(4,3) = 4 \)? But the sample space has 5 elements, and let's check which ones contain E: BCDE, ACDE, ABCE, ABDE. Wait, that's 4? But the fifth option in the problem says "There are five ways to form the committee if person E must be on it." Wait, maybe I made a mistake. Wait, the total number of 4 - member subsets of 5 elements is \( C(5,4)=5 \). Each subset either contains E or not. The number of subsets not containing E is \( C(4,4) = 1 \) (the subset ABCD). So the number of subsets containing E is \( 5-1 = 4 \)? But the sample space given is \( S=\{ABCD, BCDE, ACDE, ABCE, ABDE\} \), which are all 5 subsets. So the subsets containing E are BCDE, ACDE, ABCE, ABDE. Wait, that's 4. But the statement says 5. Wait, no, maybe I messed up. Wait, if E must be on the committee, the number of ways is the number of 4 - member committees with E. Since we need 4 members and E is one, we need 3 more from A, B, C, D. The number of ways to choose 3 from 4 is \( C(4,3)=4 \). But the sample space has 5 elements, and the one without E is ABCD, so the ones with E are 4. But the statement says 5. Wait, maybe the problem's sample space is all possible outcomes, and when E is in, let's check the sample space: BCDE (E), ACDE (E), ABCE (E), ABDE (E), and ABCD (no E). So there are 4, not 5. So the statement "There are five ways to form the committee if person E must be on it" is false? Wait, no, wait the total number of committees is 5 (from \( C(5,4)=5 \)). The number of committees with E: since each committee has 4 members, and there are 5 members, the number of committees with E is equal to the number of ways to choose 3 members from the other 4, which is \( C(4,3)=4 \). But the total number of committees is 5, so the number of committees with E is 4, not 5. So that statement is false? But wait, let's re - evaluate.

Wait, the formula for the number of subsets of size k from n elements is \( C(n,k)=\frac{n!}{k!(n - k)!} \). For \( n = 5 \), \( k = 4 \), \( C(5,4)=5 \). The number of subsets of size 4 that contain a particular element (say E) is equal to \( C(4,3) \) (because we fix E and choose 3 from the remaining 4). \( C(4,3)=\frac{4!}{3!1!}=4 \). So the statement "There are five ways to form the committee if person E must be on it" is false.

Step5: Analyze the statement "If the number of eligible members increases, the number of outcomes increases"

The number of ways to form a 4 - member committee from \( n \) eligible members is \( C(n,4)=\frac{n!}{4!(n - 4)!}=\frac{n(n - 1)(n - 2)(n - 3)}{24} \). As \( n \) increases (for \( n\geq4 \)), the value of \( C(n,4) \) increases. For example, if \( n = 6 \), \( C(6,4)=\frac{6!}{4!2!}=\frac{6\times5}{2\times1}=15 \), which is greater than \( C(5,4) = 5 \). So the statement "If the number of eligible members increases, the number of outcomes increases" is true.

Answer:

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

(The other statements are false: The total number of ways is 5, not 120; when A is on the committee, there are 4 ways, not 1; when E is on the committee, there are 4 ways, not 5)