The number of ways of distributing $n$ labelled balls amongst $k$ unlabelled boxes is $\sum_{i=0}^k{n\brace i}$, where $n\brace i$ is a Stirling number of the second kind. If we put $m$ balls into the $s$ boxes of the one type and the rest into the $k-s$ boxes of the other type, we can do this in
$$\binom{n}m\left(\sum_{i=0}^s{m\brace i}\right)\left(\sum_{i=0}^{k-s}{{n-m}\brace i}\right)=\binom{n}m\sum_{i=0}^s\sum_{j=0}^{k-s}{m\brace i}{{n-m}\brace j}\;,$$
and the desired number is therefore
$$\begin{align*}
\sum_{m=0}^n\sum_{i=0}^s\sum_{j=0}^{k-s}\binom{n}m{m\brace i}{{n-m}\brace j}&=\sum_{i=0}^s\sum_{j=0}^{k-s}\sum_{m=0}^n{m\brace i}{{n-m}\brace j}\binom{n}m\\\\
&=\sum_{i=0}^s\sum_{j=0}^{k-s}{n\brace{i+j}}\binom{i+j}i\;.\tag{1}
\end{align*}$$
In your toy example, for instance, this becomes
$$\begin{align*}
\sum_{i=0}^2\sum_{j=0}^2{3\brace{i+j}}\binom{i+j}i&=\sum_{j=0}^2{3\brace j}\binom{j}0+\sum_{j=0}^2{3\brace{j+1}}\binom{j+1}1+\sum_{j=0}^2{3\brace{j+2}}\binom{j+2}2\\
&=(0+1+3)+(1+6+3)+(3+3+0)\\
&=20\;,
\end{align*}$$
which is easily verified independently. I don’t know whether $(1)$ can be significantly simplified; I don’t see anything at the moment, but perhaps someone will have a better idea.
