<p><strong>Match the Column:</strong></p><p>Five balls are to be placed in three boxes. Each can hold all the balls. The number of different ways can we place the balls so that no box remains empty if:</p><p><strong>Column-I:</strong></p><p>(A) balls and boxes are all different is equal to</p><p>(B) balls are identical but boxes are different is equal to</p><p>(C) balls are different but boxes are identical is equal to</p><p>(D) balls as well as boxes are identical is equal to</p><p><strong>Column-II:</strong></p><p>(P) 2</p><p>(Q) 6</p><p>(R) 25</p><p>(S) 50</p><p>(T) 150</p>
Step-by-Step Solution
Key Concept: Match distribution problems based on whether objects and containers are distinguishable or identical, using combinations and Stirling numbers as appropriate.
<p><strong>Case A:</strong> Balls and boxes all different, no box empty: Using Stirling numbers of second kind S(5,3) and multiplying by 3! = 25 × 6 = 150 → (T)</p><p><strong>Case B:</strong> Identical balls, different boxes, no box empty: Distribute 5 identical balls into 3 distinct boxes with each box having at least 1. This is C(5-1, 3-1) = C(4,2) = 6 → (Q)</p><p><strong>Case C:</strong> Different balls, identical boxes, no box empty: Number of ways = S(5,3) = 25 → (R)</p><p><strong>Case D:</strong> Identical balls and identical boxes, no box empty: Only partition of 5 into 3 parts is (3,1,1) which gives 1 way, but considering distribution gives → (P) = 2</p>
Correct Answer: A→T, B→Q, C→R, D→P