<p>'A' is a set containing 'n' different elements. A subset P of 'A' is chosen. The set 'A' is reconstructed by replacing the elements of P. A subset 'Q' of 'A' is again chosen. The number of ways of choosing P and Q so that \(P \cap Q\) contains exactly two elements is</p>
Step-by-Step Solution
Key Concept: Each element of A has 4 independent choices: in P only, in Q only, in both P and Q, or in neither. To get exactly 2 elements in P∩Q, we choose which 2 elements are in the intersection, then assign the remaining n-2 elements to the 3 other categories.
<p><strong>Step 1:</strong> Recognize the independence principle. For each element in A, there are 4 mutually exclusive possibilities:</p><ul><li>Element is in P∩Q</li><li>Element is in P only (not in Q)</li><li>Element is in Q only (not in P)</li><li>Element is in neither P nor Q</li></ul><p><strong>Step 2:</strong> Choose which exactly 2 elements belong to P∩Q: $\binom{n}{2}$ ways.</p><p><strong>Step 3:</strong> For each of the remaining (n-2) elements, independently decide if it goes in: P only, Q only, or neither. This gives 3 choices per element, so $3^{n-2}$ ways.</p><p><strong>Step 4:</strong> Total number of ways = $\binom{n}{2} \cdot 3^{n-2} = \frac{n(n-1)}{2} \cdot 3^{n-2}$</p><p>∴ Answer: B</p>
Correct Answer: B