<p>Let \(S = \{1, 2, 3, 4\}\). The total number of unordered pairs of disjoint subsets of <i>S</i> is equal to</p>
Step-by-Step Solution
Key Concept: Each element of S must be assigned to exactly one of three categories: first subset, second subset, or neither. Since the pairs are unordered, we must divide by 2 to avoid double counting, except when both subsets are empty.
<p><strong>Step 1:</strong> For disjoint subsets A and B of S, each element must be in exactly one of three states: in A, in B, or in neither.</p><p><strong>Step 2:</strong> Total assignments = 3^4 = 81 (each of 4 elements has 3 choices).</p><p><strong>Step 3:</strong> Since pairs are unordered, (A, B) and (B, A) represent the same pair. We divide by 2, but must handle special cases: The case where A = B = ∅ appears once (not twice), so we subtract 1 before dividing: (81 - 1)/2 = 80/2 = 40.</p><p><strong>Step 4:</strong> We can verify: pairs with exactly k elements in A ∪ B contribute C(4,k) × 2^(k-1) unordered pairs for k ≥ 1, plus 1 for the empty pair. This gives 1 + 1 + 3 + 7 + 12 + 16 = 40.</p><p>∴ Answer: D = <strong>40</strong></p>
Correct Answer: D