Correct Option: $C$ $243$.
First take $A$ as $\emptyset$ and $B$ as power set of $\{1,2,3,4,5\} $ which is $2^5$. Then take $A$ as set of one element ex: when $A$=$\left \{ 1 \right \}$ then set $B$ could be any of the $2^4$ elements of power set. This will give us $16 \times 5$ . In similar fashion when $A$ consists of $2$ elements we get total pairs $5C2\times 2^3$; when $A$ is of three elements we get $40$; for $4$ elements we get $10$ pairs and when $A$ is of $5$ elements we get one pair which is $A$=$\left \{ 1,2,3,4,5 \right \}$ and $B=\left \{ \right \}.$ So, in total $=32+80+80+40+10+1=243$
Alternative Solution:
for each element in $[n]$, you have 3 choices:
- Include it in $A$ but not in $B$
- Include it in $B$, but not in $A $
- Include it in neither
so this gives $3^n$ pairs.
for set $\{1,2,,...,5\}$, $n=5$, $3^5=243$