1. We need the number of unordered pairs of disjoint subsets of $S=\{1,2,3,4\}.$

Let the pair be $\{A,B\}$ where $A,B\subseteq S$ and $A\cap B=\varnothing.$

2. First count ordered pairs $(A,B)$ of disjoint subsets.

For each element of $S$, there are exactly $3$ choices:

• put it in $A$,
• put it in $B$,
• put it in neither.

Since $|S|=4$, total ordered pairs are $3^4=81.$

3. Now convert ordered pairs to unordered pairs.

Usually, each unordered pair $\{A,B\}$ corresponds to two ordered pairs: $(A,B) \text{ and } (B,A).$ So we may think of dividing by $2$.

But there is one special case: when $A=B.$ Since $A\cap B=\varnothing,$ this can happen only if $A=B=\varnothing.$ So $(\varnothing,\varnothing)$ is the only ordered pair counted once under swapping.

4. Therefore, number of unordered pairs is $\frac{81-1}{2}+1 = \frac{80}{2}+1 = 40+1 = 41.$

5. Hence the correct option is $\boxed{41}.$

So, Option D is correct.