Six different balls are put in three different boxes, no box being empty. The probability of putting balls in the boxes in equal numbers is :
Step-by-Step Solution
Key Concept: Use Stirling numbers of the second kind S(n,k) to count surjective functions from n balls to k boxes. For 6 balls into 3 boxes with no box empty: total ways = 3! × S(6,3) = 6 × 90 = 540. Favorable outcomes (equal distribution 2-2-2) = 3! × (6!/(2!×2!×2!)) = 6 × 90 = 90. Probability = 90/540 = 1/6.
Total ways to distribute balls such that no box is empty uses surjective functions: $\frac{3!}{2!}\left[\binom{6}{1} \cdot \binom{1}{1} \cdot \binom{1}{1}\right] + 3!\left[\binom{2}{2} \cdot \binom{1}{1} \cdot \binom{1}{1}\right] + \binom{2}{2} \cdot \binom{2}{2} \cdot \binom{2}{2} = 90 + 6.60 + 90 = 540$. Required probability is $\frac{90}{540} = \frac{1}{6}$.
Correct Answer: 2