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(D) For each element in the set $S$,there are $3$ possibilities regarding its membership in two disjoint subsets $A$ and $B$: $1$. The element is in $

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$41$

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(D) For each element in the set $S$,there are $3$ possibilities regarding its membership in two disjoint subsets $A$ and $B$: $1$. The element is in $A$ but not in $B$. $2$. The element is in $B$ but not in $A$. $3$. The element is in neither $A$ nor $B$. Since there are $n = 4$ elements,the total number of ordered pairs of disjoint subsets $(A, B)$ is $3^n = 3^4 = 81$. To find the number of unordered pairs ${A, B}$,we must account for the case where $A = B$. Since $A$ and $B$ must be disjoint,$A = B$ implies $A = B = \emptyset$. This occurs in exactly $1$ case. For all other cases where $A 
eq B$,the pair ${A, B}$ is counted twice in the ordered list (as $(A, B)$ and $(B, A)$). The number of unordered pairs is given by $\frac{3^n + 1}{2} = \frac{3^4 + 1}{2} = \frac{81 + 1}{2} = \frac{82}{2} = 41$.

Let $S = \{1, 2, 3, 4\}$. The total number of unordered pairs of disjoint subsets of $S$ is equal to

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