Sets, Relations & Functions
Sets and their properties
Grade 11
Question:
<p>Suppose <em>A</em><sub>1</sub>, <em>A</em><sub>2</sub>, ..., <em>A</em><sub>30</sub> are thirty sets each having 5 elements and <em>B</em><sub>1</sub>, <em>B</em><sub>2</sub>, ..., <em>B</em><sub>n</sub> are <em>n</em> sets each having 3 elements. Let \(\bigcup_{i=1}^{30} A_i = \bigcup_{j=1}^{n} B_j = S\) and each element of <em>S</em> belongs to exactly 10 of the <em>A</em><sub>i</sub>'s and exactly 9 of the <em>B</em><sub>j</sub>'s, then find the value of <em>n</em>.</p>
Step-by-Step Solution
Key Concept: Use the counting principle: sum of cardinalities equals cardinality of union times average frequency. Since each element of S appears in exactly 10 A's and exactly 9 B's, equate |A₁| + |A₂| + ... + |A₃₀| to 10|S| and |B₁| + |B₂| + ... + |Bₙ| to 9|S|.
<p><strong>Step 1:</strong> Count total elements in all Aᵢ's with multiplicity.</p><p>Since each Aᵢ has 5 elements: ∑|Aᵢ| = 30 × 5 = 150</p><p><strong>Step 2:</strong> Relate this sum to S and frequency.</p><p>Each element of S belongs to exactly 10 of the Aᵢ's, so when we sum all |Aᵢ|, each element of S is counted 10 times.</p><p>Therefore: 10|S| = 150, which gives |S| = 15</p><p><strong>Step 3:</strong> Count total elements in all Bⱼ's with multiplicity.</p><p>Since each Bⱼ has 3 elements: ∑|Bⱼ| = n × 3 = 3n</p><p><strong>Step 4:</strong> Relate this sum to S and frequency.</p><p>Each element of S belongs to exactly 9 of the Bⱼ's, so each element of S is counted 9 times.</p><p>Therefore: 9|S| = 3n</p><p><strong>Step 5:</strong> Solve for n.</p><p>9 × 15 = 3n</p><p>135 = 3n</p><p>∴ n = 45</p>
Correct Answer: 45