Sets, Relations & Functions
Symmetric Difference of Sets
Grade 11
Question:
<p>Let <em>A</em>, <em>B</em> and <em>C</em> be finite sets such that \(A \cap B \cap C = \phi\) and each one of the sets \(A \Delta B\), \(B \Delta C\) and \(C \Delta A\) has 100 elements. Given that \(A \Delta B = (A \cup B) - (A \cap B)\), the number of elements in \(A \cup B \cup C\) is __________.</p>
Step-by-Step Solution
Key Concept: Since A ∩ B ∩ C = ∅, the symmetric differences A △ B, B △ C, C △ A partition into disjoint regions. Using the property that each symmetric difference equals 100, we can express |A ∪ B ∪ C| by counting overlaps systematically: each pairwise intersection (excluding the triple intersection) contributes to exactly two symmetric differences.
<p><strong>Step 1:</strong> Define regions. Let:</p><ul><li>x = |A ∩ B - C| (in A and B, not C)</li><li>y = |B ∩ C - A| (in B and C, not A)</li><li>z = |C ∩ A - B| (in C and A, not B)</li><li>a = |A - (B ∪ C)|, b = |B - (A ∪ C)|, c = |C - (A ∪ B)| (elements only in one set)</li></ul><p><strong>Step 2:</strong> Express symmetric differences using these regions:</p><ul><li>A △ B = (A - B) ∪ (B - A) = (a + z) + (b + x) = a + b + x + z = 100</li><li>B △ C = (B - C) ∪ (C - B) = (b + x) + (c + y) = b + c + x + y = 100</li><li>C △ A = (C - A) ∪ (A - C) = (c + y) + (a + z) = a + c + y + z = 100</li></ul><p><strong>Step 3:</strong> Add all three equations:</p><p>(a + b + x + z) + (b + c + x + y) + (a + c + y + z) = 300</p><p>2a + 2b + 2c + 2x + 2y + 2z = 300</p><p>a + b + c + x + y + z = 150</p><p><strong>Step 4:</strong> Recognize that |A ∪ B ∪ C| = a + b + c + x + y + z (since A ∩ B ∩ C = ∅, all elements are accounted for by these seven disjoint regions)</p><p>∴ Answer: <strong>150</strong></p>
Correct Answer: 150