Let S be a set of 5 elements and P(S) denote | vedclass

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(C) Let $S = \{a, b, c, d, e\}$ be a set with $n = 5$ elements.For any element $x \in S$, there are $4$ possibilities for its membership in the sets $

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) Let $S = \{a, b, c, d, e\}$ be a set with $n = 5$ elements.For any element $x \in S$, there are $4$ possibilities for its membership in the sets $A$ and $B$ such that $A \cap B = \varnothing$:$1$. $x \in A$ and $x 
otin B$$2$. $x 
otin A$ and $x \in B$$3$. $x 
otin A$ and $x 
otin B$(Note: $x \in A$ and $x \in B$ is not allowed as $A \cap B = \varnothing$)Since there are $5$ elements, the total number of ordered pairs $(A, B)$ is $(2^n) 	imes (2^n) = 2^5 	imes 2^5 = 2^{10} = 4^5$.The number of favourable pairs $(A, B)$ such that $A \cap B = \varnothing$ is $3^5$ (since each element has $3$ choices).Thus, the probability $P(E) = \frac{3^5}{4^5} = \frac{3^5}{(2^2)^5} = \frac{3^5}{2^{10}}$.Comparing this with $\frac{3^p}{2^q}$, we get $p = 5$ and $q = 10$.Therefore, $p + q = 5 + 10 = 15$.

Let $S$ be a set of $5$ elements and $P(S)$ denote the power set of $S$. Let $E$ be an event of choosing an ordered pair $(A, B)$ from the set $P(S) \times P(S)$ such that $A \cap B = \varnothing$. If the probability of the event $E$ is $\frac{3^p}{2^q}$, where $p, q \in N$, then $p+q$ is equal to

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