<s>[INST] Probability [Distinct balls in distinct boxes] 6 different balls are put in 3 different boxes, no box being empty. The probability of putting balls in the boxes in equal numbers is ? Could anyone pleases give the answer and explain it.
 [/INST] Note that the balls and boxes are both distinct.
Select 2 balls in each of 3 boxes, count the permutations of the six balls. Ways to arrange 6 balls, 2 in each box: $N(2,2,2)=\frac{6!}{2!2!2!}$  That's permutations of all six balls, divided by permutations of balls in each box (since the order in the box does not matter).
Select the 1 box to have 4 balls, permute the balls.  Ways to arrange 6 balls, 1 in each of 2 boxes and 4 in the third: $N(1,1,4) = \frac{6!}{4!}\times {3 \choose 1} = \frac{6!}{4!}\times 3$
Select 1 of 3 boxes to have 3 balls, select 1 of remaining 2 to have 2 balls, permute the balls. Ways to arrange 6 balls, 1 in one box, 2 in another, 3 into the third: $N(1,2,3) = \frac{6!}{2!3!}\times{3 \choose 1}{2 \choose 1} = \frac{6!}{2!3!} \times 3 \times 2$
Given that no box can be empty, that is all the ways to fill the boxes.  So we want the probability : $$\dfrac{N(2,2,2)}{N(2,2,2)+N(1,1,4)+N(1,2,3)} \\ = \dfrac{\frac{1}{2!2!2!}}{\frac{1}{2!2!2!}+\frac{3}{4!}+\frac{1}{3!2!}\cdot 3\cdot 2} \\ = \frac{4!}{4!+2!2!2!3+2!2!4!} \\ = \dfrac{1}{6}$$
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