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

$\text{Given: } S = \{1, 2, 3, 4\}$

$\text{To find: number of unordered pairs } \{A, B\} \text{ of subsets of } S$

$\text{such that } A \cap B = \varnothing$

$\text{Unordered pair: } \{A, B\} = \{B, A\}$

$\text{Disjoint: } A \cap B = \varnothing$

$\text{Plan: count ordered pairs first, then remove the order}$

$\text{Each } x \in S \text{ has exactly 3 choices:}$

$x \in A \quad \text{or} \quad x \in B \quad \text{or} \quad x \notin A \cup B$

$\text{4 independent choices} \implies 3 \times 3 \times 3 \times 3 = 3^4$

$3^4 = 81 \quad \text{ordered pairs } (A, B)$

$\text{Ordered count counts } (A, B) \text{ and } (B, A) \text{ separately}$

$\text{Swap } (A, B) \leftrightarrow (B, A) \text{ pairs them up}$

$(A, B) = (B, A) \iff A = B$

$A \cap B = \varnothing \implies A = B = \varnothing$

$\text{Unordered} = \dfrac{81 - 1}{2} + 1$

$= \dfrac{80}{2} + 1 = 40 + 1$

$\text{General: unordered disjoint pairs} = \dfrac{3^n + 1}{2}$

$\text{Check } n = 1:\ \dfrac{3 + 1}{2} = 2$

$\sum_{k=0}^{4} \binom{4}{k} 2^{4-k} = (1 + 2)^4 = 81$

$n = 4:\ \dfrac{3^4 + 1}{2} = \dfrac{82}{2} = 41$

$\text{Answer: } \boxed{41} \quad (\text{option } 4)$

$\text{Key idea: 3 choices per element}$

$\text{then } \dfrac{3^n - 1}{2} + 1 = \dfrac{3^n + 1}{2}$