Sets, Relations & Functions
Set operations and counting
Grade 11
<p>Let <i>X</i> = {1, 2, 3, 4, 5}. The number of different ordered pairs (<i>Y</i>, <i>Z</i>) that can be formed such that <i>Y</i> ⊆ <i>X</i>, <i>Z</i> ⊆ <i>X</i> and <i>Y</i> ∩ <i>Z</i> is empty, is</p>
Step-by-Step Solution
Key Concept: For each element in X, we have exactly 3 independent choices: put it in Y only, put it in Z only, or put it in neither (since Y ∩ Z = ∅ means no element can be in both). This gives 3^|X| total ordered pairs.
<p><strong>Step 1:</strong> Understand the constraint. We need ordered pairs (Y, Z) where both Y and Z are subsets of X = {1, 2, 3, 4, 5}, and Y ∩ Z = ∅ (disjoint).</p><p><strong>Step 2:</strong> For each element in X, determine its possible assignments. Since Y and Z must be disjoint, each element has exactly 3 mutually exclusive options:</p><ul><li>Element belongs to Y only</li><li>Element belongs to Z only</li><li>Element belongs to neither Y nor Z</li></ul><p><strong>Step 3:</strong> Apply the multiplication principle. Since we have 5 elements and each has 3 independent choices, the total number of ordered pairs (Y, Z) is:</p><p>3 × 3 × 3 × 3 × 3 = 3^5 = 243</p><p><strong>Step 4:</strong> Verify with small case: For X = {1}, we get pairs: ({∅}, {∅}), ({1}, {∅}), ({∅}, {1}) = 3 pairs ✓</p><p>∴ Answer: <strong>243</strong></p>
Correct Answer: B