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.\nthere are 120 different ways to choose the committee.\nif person a must be on the committee, there is only one way to form the committee.\nif persons a and c must be on the committee, there are three ways to form the committee.\nthere are five ways to form the committee if person e must be on it.\nif 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.\nthere are 120 different ways to choose the committee.\nif person a must be on the committee, there is only one way to form the committee.\nif persons a and c must be on the committee, there are three ways to form the committee.\nthere are five ways to form the committee if person e must be on it.\nif the number of eligible members increases, the number of outcomes increases.

Answer

Explanation:

Step1: Calcular el número total de combinaciones

Usamos la fórmula de combinaciones $C(n,r)=\frac{n!}{r!(n - r)!}$, donde $n = 5$ (miembros elegibles) y $r=4$ (miembros del comité). Entonces $C(5,4)=\frac{5!}{4!(5 - 4)!}=\frac{5!}{4!1!}=\frac{5\times4!}{4!}=5$.

Step2: Analizar el caso de A en el comité

Si A debe estar en el comité, entonces debemos elegir 3 miembros restantes de 4. $C(4,3)=\frac{4!}{3!(4 - 3)!}=\frac{4!}{3!1!}=4$.

Step3: Analizar el caso de A y C en el comité

Si A y C deben estar en el comité, entonces debemos elegir 2 miembros restantes de 3. $C(3,2)=\frac{3!}{2!(3 - 2)!}=\frac{3!}{2!1!}=3$.

Step4: Analizar el caso de E en el comité

Si E debe estar en el comité, entonces debemos elegir 3 miembros restantes de 4. $C(4,3)=\frac{4!}{3!(4 - 3)!}=\frac{4!}{3!1!}=4$.

Step5: Analizar el aumento del número de miembros elegibles

Según la fórmula de combinaciones, si $n$ aumenta (mientras $r$ permanece constante), el valor de $C(n,r)$ aumentará.

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.