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: The probability requires counting surjective functions (onto mappings) where all 6 distinct balls go to 3 distinct boxes with no box empty. For equal distribution (2 balls per box), use multinomial coefficients: favorable outcomes = 3! × C(6,2) × C(4,2) × C(2,2) = 90, and total surjective functions = 540, giving 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