<p>Let \(A\) and \(B\) be two sets containing four and two elements respectively. Then the number of subsets for the set \(A \times B\), each having at least three elements is</p>
Step-by-Step Solution
Key Concept: The Cartesian product A × B has |A| × |B| = 4 × 2 = 8 elements total. We need subsets with at least 3 elements, so count subsets of sizes 3, 4, 5, 6, 7, and 8 using combinations: C(8,3) + C(8,4) + C(8,5) + C(8,6) + C(8,7) + C(8,8).
<p><strong>Step 1:</strong> Find the cardinality of A × B.</p><p>Since |A| = 4 and |B| = 2, we have |A × B| = 4 × 2 = 8 elements.</p><p><strong>Step 2:</strong> Count subsets of A × B with at least 3 elements.</p><p>Total subsets with at least 3 elements = C(8,3) + C(8,4) + C(8,5) + C(8,6) + C(8,7) + C(8,8)</p><p><strong>Step 3:</strong> Calculate each combination.</p><p>C(8,3) = 56<br>C(8,4) = 70<br>C(8,5) = 56<br>C(8,6) = 28<br>C(8,7) = 8<br>C(8,8) = 1</p><p><strong>Step 4:</strong> Sum the values.</p><p>Total = 56 + 70 + 56 + 28 + 8 + 1 = 219</p><p><strong>Alternative:</strong> Total subsets = 2^8 = 256. Subtract subsets with 0, 1, or 2 elements: 256 − [C(8,0) + C(8,1) + C(8,2)] = 256 − [1 + 8 + 28] = 256 − 37 = 219</p><p>∴ Answer: D (219)</p>
Correct Answer: D