Balls and Boxes gone crazy. Part-II

4 gaps exist between 5 balls in a row,

To separate them into 3 groups (Right, Center, Left)

We choose 2 gaps to place our separator.

make it $4C2=6$

There are 3 different amounts of balls one can put in one box: 1, 2, and 3.

(This is because as the problem itself stated, no box remains empty, so the amount cannot be 0. Also it cannot be 4 or 5, because they imply that at least one box is empty.)

Therefore, there are 3! ways of arranging the balls.

The balls could be placed in three boxes in the permutations of the configuration $\{3, 1, 1\}$ or $\{2, 2, 1\}\ = 3 + 3 = 6$ .

$\{3, 1, 1\}$ means $1^{st}$ box gets $3$ balls, and, $2^{nd}$ and $3^{rd}$ box gets $1$ ball each.

$\{2, 2, 1\}$ means $1^{st}$ and $2^{nd}$ box gets $2$ balls each, and $3^{rd}$ box gets $1$ ball.

Since all boxes must have at least one ball in it, we can start with one ball in each box. This leaves us with 2 balls left, we can thus rephrase the question: in how many ways can we allocate 2 objects in 3 locations with repetition? That's 6.

The question is similar to finding the positive integral solutions of the equation a+b+c=5 which is simply 4C2 = 6