Probability
Probability
Allen Star Batch
Grade 12
Question:
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 :
$3/10$
$1/6$
$1/5$
None of these
Step-by-Step Solution
Key Concept: Count surjective distributions (no empty boxes) by partitioning ball distributions among boxes.
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