Sets, Relations & Functions
Operations on Sets
Grade 11
Question:
<p>Let <em>X</em> = {1, 2, 3, 4, 5}. The number of different ordered pairs (<em>Y</em>, <em>Z</em>) that can be formed such that \(Y \subseteq X\), \(Z \subseteq X\) and \(Y \cap Z\) is empty, is</p>
<p>\(5^2\)</p>
<p>\(3^5\)</p>
<p>\(2^5\)</p>
<p>\(5^3\)</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 (we cannot put it in both since Y ∩ Z = ∅). This gives 3^|X| total ordered pairs.
<p><strong>Step 1:</strong> Understand the constraint. We need Y ⊆ X, Z ⊆ X, and Y ∩ Z = ∅ (Y and Z are disjoint).</p><p><strong>Step 2:</strong> For each element x ∈ X, determine its possibilities:</p><ul><li>x ∈ Y and x ∉ Z</li><li>x ∉ Y and x ∈ Z</li><li>x ∉ Y and x ∉ Z</li><li>x ∉ Y and x ∉ Z (cannot have x in both since Y ∩ Z = ∅)</li></ul><p><strong>Step 3:</strong> Simplify: For each element, there are exactly 3 independent choices:</p><ul><li>Element goes to Y only</li><li>Element goes to Z only</li><li>Element goes to neither</li></ul><p><strong>Step 4:</strong> Since |X| = 5 and each of the 5 elements has 3 independent choices, the total number of ordered pairs (Y, Z) is:</p><p>3 × 3 × 3 × 3 × 3 = <strong>3^5 = 243</strong></p><p>∴ Answer: B</p>
Correct Answer: B