Find the number of distributions of $5$ distinct balls into $2$ red boxes and $1$ blue box if the $2$ red boxes are identical and no boxes are empty.

Question

$S(m, n)$ stands for Stirling numbers of second kind.

Find the number of distributions of $5$ distinct balls into $2$ red boxes and $1$ blue box if the $2$ red boxes are identical and no boxes are empty.

I think there are at least two ways to go about this problem

• Pretend that the red boxes are distinct. Then by twelvefold way, there are $3! \cdot S(5, 3)$ distributions of $5$ distinct balls to $3$ distinct boxes. But since the red balls are actually identical we overcounted by a factor of $2!$ Thus the answer is $\frac{3! \cdot S(5, 3)}{2!} = 75.$

• Pretend all the boxes are identical. Then we have $S(5, 3)$ partitions of $5$ distinct balls to $3$ identical boxes. Suppose now we line up the boxes. Then the blue box can be in the first position or in the middle of two red boxes or in the third position. So the answer is $3 \cdot S(5, 3) = 75.$

Does that above-written look correct to you?

I don't want to clutter up this site with trivial questions so I decided to ask a very quick add-on question:

Find the number of ways to distribute five distinct balls into three identical boxes if

(a) no boxes are empty;

(b) empty boxes are allowed;

(c) at most one box is empty.

Do the solutions below work?

(a) $S(5, 3)$

(b) $S(5, 2) + S(5, 1)$

(c) $S(5, 2)$

Thanks.