The total number of ways of distributing 12 identical balls into 3 DISTINCT boxes would be (12+3-1)c(3-1) = 91

Now, as the boxes are identical, what would be the difference? If I have a distribution of 1,2,9 balls in the boxes, then this specific distribution can be done 3!=6 ways when the boxes are distinct, but in only ONE way, if they are identical.

Similarly, if we have a distribution of the type 2,2,8 balls, then this can be done in 3!/2!=3 ways in distinct boxes, but again in only ONE way for identical boxes.

If, we have the distribution 4,4,4 balls, then that can be done in only ONE way in both the cases.

So, first I'll distribute the 91 cases into three types:

All distinct number of ballsTwo identical number of ballsAll identical number of balls

What we are going to do is, we'll begin with type 2. In this scenario, we have let's say a,a,b balls in the boxes.

We have 2a+b=12, whose number of solutions is 6+1=7

But out of those 7 cases, 1 is the distribution 4,4,4 which does not fall under this scenario.

So, we have 6 cases under type 3, each of which has been counted thrice, and 1 case of type 3, which has been counted once. And all other are of type 1 which has counted 6 times

Hence distinct possibilities are = 6/3 + 1 + (91-7)/6 = 17

This would determine all possible outcomes.

Now, it's easy to find the favourable outcomes

3+a+b=12

Number of solutions for (a,b) will be (0,9); (1,8) ... (4,5): which is a total of 5 cases

So, probability = 5/17