<p>Let \(S = \{1, 2, 3, \ldots, 100\}\). The number of non-empty subsets \(A\) of \(S\) such that the product of elements in \(A\) is even is:</p>
Step-by-Step Solution
Key Concept: A product is even if and only if at least one element in the set is even. Count subsets with at least one even number by using complementary counting: total non-empty subsets minus subsets containing only odd numbers.
<p><strong>Step 1:</strong> Identify the structure of S. In S = {1, 2, 3, ..., 100}, there are 50 odd numbers {1, 3, 5, ..., 99} and 50 even numbers {2, 4, 6, ..., 100}.</p><p><strong>Step 2:</strong> Recognize that a product is even ⟺ at least one element is even. Use complementary counting: subsets with even product = total non-empty subsets − non-empty subsets with only odd numbers.</p><p><strong>Step 3:</strong> Calculate total non-empty subsets of S: 2^100 − 1</p><p><strong>Step 4:</strong> Calculate non-empty subsets of only odd numbers: The 50 odd numbers form a set with 2^50 − 1 non-empty subsets.</p><p><strong>Step 5:</strong> Apply complementary counting: Number of subsets with even product = (2^100 − 1) − (2^50 − 1) = 2^100 − 2^50</p><p><strong>Step 6:</strong> Factor: 2^100 − 2^50 = 2^50(2^50 − 1)</p><p>∴ Answer: <strong>2^100 − 2^50</strong> or equivalently <strong>2^50(2^50 − 1)</strong></p>
Correct Answer: B