Number of ways in which 3 identical red balls and 3 identical white balls can be distributed in between 3 distinct boxes, no box is empty? As mentioned in the title, we need to calculate the number of ways in which 3 identical red balls and 3 identical white balls can be distributed in between 3 distinct boxes such that no box is empty.
There have been a few similar questions asked, but none which completely answers this particular question (as per my knowledge).
I tried to approach this by making a few cases, which actually ended up working. But I was not able to create a general approach for, say n identical objects of one type and m identical objects of another type in p different boxes.
 A: At the start we have $6$ white balls. We can have  $\{4,1,1\}$, $\{3,2,1\}$, or $\{2,2,2\}$ balls in the boxes, with $3$, $6$, $1$ different orderings in the three cases. We now paint three of the six balls red. In the $\{4,1,1\}$ case we can paint three of the $4$ red ($1$ way), two of the $4$ red ($2$ ways), or one of the $4$ red ($1$ way); makes $4$ ways. In the $\{3,2,1\}$ case we can paint all three of the $3$ red ($1$ way), two of the three red ($2$ ways), one of the $3$ red ($2$ ways), or none of the $3$ red ($1$ way); makes $6$ ways. In the $\{2,2,2\}$ case we can make $2$ and $1$ red balls in different boxes ($6$ ways) or one red ball in each box ($1$ way); makes $7$ ways.
In all, there are
$$3\cdot 4+6\cdot 6+1\cdot 7=55$$
different admissible distributions.
A: Case A. 4 balls in the first box.

*

*In the box we can find 3 red balls & 1 white or 3 white balls & 1 red. This means exactly one arranjament for second and third box. Subtotal:2 permutations

*In the box we can find 2 red balls & 2 white balls. This means two possivle arranjaments for second and third box. Subtotal:2 permutations
Total:4 permutations
Case B. 3 balls in the first box.

*

*3 red or 3 white. This means 2 arranjaments in the other boxes. Subtotal:4 permutations

*2 red + 1 white or 1 red + 2 white. This means 4 possible arranjaments in the other boxes. Subtotal:8 permutations
Total:12 permutations
Case C. 2 balls in the first box.

*

*2 red or 2 white. This means 6 possible arranjaments in the other boxes. Subtotal:12 permutations

*1 red & 1 white. This means 7 possible arranjaments in the other boxes. Subtotal:7 permutations
Total:19 permutations
Case D. 1 ball in the first box.
Only one way: 1 red or 1 white. This means 10 possible arranjaments in the other boxes.
Total:20 permutations
Conclusion: 4 + 12 + 19 + 20 = 55 possible permutations.
