Let A = {a, b, c, d, e} and P(A) = S where P(A) is the power set of A. Number of elements (P, Q) in S × S such that P ∩ Q = φ is equal to

The correct answer is 243.

Let the given set be
$A=\{a,b,c,d,e\}$
containing 5 elements. The power set of $A$ is denoted as $S=P\left(A\right)$.

We need to find the number of ordered pairs
$(P,Q)$
in $S\times S$ such that
$P\cap Q=\varnothing$.

For each element in the set $A$, say $x\in A$, there are three mutually exclusive possibilities to form the subsets $P$ and $Q$ such that their intersection is empty:
1. $x\in P$ and $x\notin Q$
2. $x\notin P$ and $x\in Q$
3. $x\notin P$ and $x\notin Q$

Since the set $A$ contains 5 distinct elements, and each element has exactly 3 choices, the total number of such ordered pairs is:

$3^5=243$