4. Si $$ \mathrm{A} \subset B \subset C $$ $$ n(B)=n(A)+5 $$$$ \quad \begin{array}{ll}n(C)=2 imes n(B) \ n(A)+n(B)+n(C)=27 \ ext { halla } n[P(C-B)] . & \ \begin{array}{lll} ext { A) } 48 & ext { B) } 8 & ext { C) } 256 \ ext { D) } 16 & ext { E) } 32 & \end{array}\end{array} $$

Question

4. Si $$ \mathrm{A} \subset B \subset C $$ $$ n(B)=n(A)+5 $$$$ \quad \begin{array}{ll}n(C)=2 	imes n(B) \ n(A)+n(B)+n(C)=27 \ 	ext { halla } n[P(C-B)] . & \ \begin{array}{lll}	ext { A) } 48 & 	ext { B) } 8 & 	ext { C) } 256 \ 	ext { D) } 16 & 	ext { E) } 32 & \end{array}\end{array} $$

UpStudy AI · Accepted Answer

1. Planteamos las relaciones dadas:
   
   $$
   n(B) = n(A) + 5
   $$
   $$
   n(C) = 2 \times n(B)
   $$
   $$
   n(A)+n(B)+n(C) = 27
   $$

2. Sea $$ n(A) = a $$. Entonces:

$$
   n(B) = a + 5 \quad \text{y} \quad n(C) = 2(a+5)
   $$

3. Sustituimos en la suma total:

$$
   a + (a+5) + 2(a+5) = 27
   $$
   
   Expandimos y simplificamos:

$$
   a + a + 5 + 2a + 10 = 27
   $$
   
   $$
   4a + 15 = 27
   $$

4. Resolviendo para $$ a $$:

$$
   4a = 27 - 15 = 12 \quad \Longrightarrow \quad a = 3
   $$
   
   Por lo tanto:

$$
   n(A) = 3 \quad,\quad n(B) = 3 + 5 = 8 \quad,\quad n(C) = 2 \times 8 = 16
   $$

5. Se pide hallar el número de subconjuntos del conjunto $$ C-B $$. Calculamos:

$$
   n(C-B) = n(C) - n(B) = 16 - 8 = 8
   $$

6. El número de subconjuntos del conjunto $$ C-B $$ es:

$$
   n[P(C-B)] = 2^{n(C-B)} = 2^8 = 256
   $$

La respuesta correcta es 256.
$$ n[P(C-B)] = 256 $$, opción C.

Si

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