 Solution

For one colour, the box occupancies form a weak composition of $k$ into $n$ nonnegative parts. For $n\ge 1$, stars and bars gives $\left(\genfrac{}{}{0ex}{}{k+n-1}{n-1}\right)$ choices. The three colours' occupancies can be selected independently, so the number isBoxes are labelled, while equal-colour balls contribute only occupancy counts; no extra factorial for permuting such balls is appropriate.