<p>Five different marbles are placed in 5 different boxes randomly. Then the probability that exactly two boxes remain empty is (each box can hold any number of marbles)</p>
Step-by-Step Solution
Key Concept: Use Stirling numbers of the second kind S(n,k) to count surjections: we need exactly 3 boxes filled (5 marbles onto 3 boxes, each box gets ≥1 marble), then multiply by ways to choose which 3 boxes. The probability is [C(5,3)·S(5,3)]/5^5.
<p><strong>Step 1:</strong> If exactly 2 boxes are empty, then exactly 3 boxes are non-empty. We need to distribute 5 distinguishable marbles into exactly 3 distinguishable boxes such that each box has at least one marble (surjection).</p><p><strong>Step 2:</strong> The number of surjections from a set of 5 elements onto a set of 3 elements is given by Stirling number of second kind: S(5,3)·3! = 25·6 = 150. Alternatively, using inclusion-exclusion: 3^5 - C(3,1)·2^5 + C(3,2)·1^5 = 243 - 96 + 3 = 150.</p><p><strong>Step 3:</strong> We must choose which 3 boxes (out of 5) remain non-empty: C(5,3) = 10 ways.</p><p><strong>Step 4:</strong> Total favorable outcomes = 10 × 150 = 1500. [Note: Here boxes and marbles are both distinguishable, so we count ordered distributions directly]</p><p><strong>Step 5:</strong> Actually, more directly: number of ways to place 5 distinguishable marbles into 5 distinguishable boxes with exactly 3 non-empty = C(5,3) × S(5,3) × 3! where S(5,3)=25. But the surjection count already includes ordering, so: C(5,3) × [3^5 - C(3,1)·2^5 + C(3,2)·1] = 10 × 150 = 1500.</p><p><strong>Step 6:</strong> Total possible ways = 5^5 = 3125.</p><p><strong>Step 7:</strong> Probability = 1500/3125 = 12/25.</p><p>∴ Answer: 12/25</p>
Correct Answer: 2