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 200 elements. The number of elements in \(A \cup B \cup C\) is __________.</p>
Step-by-Step Solution
Key Concept: Use the property that A△B = (A−B)∪(B−A) and the constraint that A∩B∩C = ∅ means no element belongs to all three sets simultaneously. Express |A∪B∪C| in terms of elements in exactly one or two sets using symmetric difference equations.
<p><strong>Step 1:</strong> Since A∩B∩C = ∅, partition A∪B∪C into disjoint regions:</p><ul><li>Elements only in A: a</li><li>Elements only in B: b</li><li>Elements only in C: c</li><li>Elements in A∩B but not C: x</li><li>Elements in B∩C but not A: y</li><li>Elements in C∩A but not B: z</li></ul><p><strong>Step 2:</strong> Express symmetric differences using the partition:</p><ul><li>A△B = (A−B)∪(B−A) = (a+z)+(b+y) = a+b+y+z = 200</li><li>B△C = (B−C)∪(C−B) = (b+x)+(c+z) = b+c+x+z = 200</li><li>C△A = (C−A)∪(A−C) = (c+y)+(a+x) = a+c+x+y = 200</li></ul><p><strong>Step 3:</strong> Add all three equations:</p><p>(a+b+y+z) + (b+c+x+z) + (a+c+x+y) = 600</p><p>2a + 2b + 2c + 2x + 2y + 2z = 600</p><p>a + b + c + x + y + z = 300</p><p><strong>Step 4:</strong> Since |A∪B∪C| = a + b + c + x + y + z</p><p>∴ Answer: <strong>300</strong></p>
Correct Answer: 300