The answer given here is actually to be considered an addendum to
the answer given to the other post, where it has been considered the number of ways $C(m,n,l)$ to arrange:
- $l$ undistinguishable balls
- into $m$ distinguishable boxes, each provided with $n$ distinguishable compartments
- counting only the arrangements that contain no empty boxes = at least $1$ ball per box.
Let's now call $C_{2}(m,n,l)$ the number of ways to arrange balls in boxes same as above, but
- counting only the arrangements that contain at least $2$ balls per box.
It is evident that the approach described in the previous answer and leading to
the involutory recurrence 4) therein, can be applied here as well.
Briefly,
- imagine to have the sketch of all the configurations pertaining to $C_{2}(m,n,l)$;
- add an additional box aside;
- transfer one ball at time from the original block to the new box;
- starting from the 2nd ball, you are constructing a partition of the $(m+1, \, n,\, l)$ arrangement,
that at each step accounts for $C_{2}(m,n,l-k) \cdot C_{2}(1,n,k)$;
- the process stops when $k=n$ or $k=l$, but we can sum for $0 \leqslant k \leqslant l$
because by definition $C_{2}(m,n,j) = 0$ for $j \leq 2m$ .
Thus we have
$$
C_{\,2} (m + 1,n,l)\quad \left| {\;0 \leqslant \text{integer}\;m,n,l} \right.\quad =
\sum\limits_{\left( {0\, \leqslant } \right)\;k\,\,\left( { \leqslant \,l} \right)}
{C_{\,2} (m,n,l - k)\;C_{\,2} (1,n,k)} \tag {4.a}
$$
that is the same recurrence as in previous answer.
But now the initial conditions will be
$$
C_{\,2} (1,n,l)\quad = \quad \begin{array}{*{20}c}
{n\backslash l} &| & 0 & 1 & {2 \leqslant l} \\
\hline
0 &| & 0 & 0 & 0 \\
1 &| & 0 & 0 & 0 \\
{2 \leqslant n} &| & 0 & 0 & {\left( \begin{gathered}
n \\
l \\
\end{gathered} \right)} \\
\end{array} \quad \quad = \quad \left( \begin{gathered}
n \\
l \\
\end{gathered} \right) - \left( \begin{gathered}
0 \\
l \\
\end{gathered} \right) - n\left( \begin{gathered}
0 \\
l - 1 \\
\end{gathered} \right)
$$
so, we have
$$
\begin{gathered}
C_{\,2} (2,n,l) = \sum\limits_{\left( {0\, \leqslant } \right)\;k\,\,\left( { \leqslant \,l} \right)} {\left( {\left( \begin{gathered}
n \\
l - k \\
\end{gathered} \right) - \left( \begin{gathered}
0 \\
l - k \\
\end{gathered} \right) - n\left( \begin{gathered}
0 \\
l - k - 1 \\
\end{gathered} \right)} \right)\;\left( {\left( \begin{gathered}
n \\
k \\
\end{gathered} \right) - \left( \begin{gathered}
0 \\
k \\
\end{gathered} \right) - n\left( \begin{gathered}
0 \\
k - 1 \\
\end{gathered} \right)} \right)} = \hfill \\
= \left( \begin{gathered}
2n \\
l \\
\end{gathered} \right) - \left( \begin{gathered}
n \\
l \\
\end{gathered} \right) - \left( \begin{gathered}
n \\
l - 1 \\
\end{gathered} \right) - \left( \begin{gathered}
n \\
l \\
\end{gathered} \right) + \left( \begin{gathered}
0 \\
l \\
\end{gathered} \right) + n\left( \begin{gathered}
0 \\
l - 1 \\
\end{gathered} \right) - n\left( \begin{gathered}
n \\
l - 1 \\
\end{gathered} \right) + n\left( \begin{gathered}
0 \\
l - 1 \\
\end{gathered} \right) + n^{\,2} \left( \begin{gathered}
0 \\
l - 2 \\
\end{gathered} \right) = \hfill \\
= \left( {\left( \begin{gathered}
2n \\
l \\
\end{gathered} \right) - 2\left( \begin{gathered}
n \\
l \\
\end{gathered} \right) + \left( \begin{gathered}
0 \\
l \\
\end{gathered} \right)} \right) - 2n\left( {\left( \begin{gathered}
n \\
l - 1 \\
\end{gathered} \right) - \left( \begin{gathered}
0 \\
l - 1 \\
\end{gathered} \right)} \right) + n^{\,2} \left( \begin{gathered}
0 \\
l - 2 \\
\end{gathered} \right) = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\;j\,\,\left( { \leqslant \,2} \right)} {\left( { - 1} \right)^{\,j} \left( \begin{gathered}
2 \\
j \\
\end{gathered} \right)\left( {\sum\limits_{\left( {0\, \leqslant } \right)\;k\,\,\left( { \leqslant \,2} \right)} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
2 - j \\
k \\
\end{gathered} \right)\left( \begin{gathered}
\left( {2 - j - k} \right)n \\
l - j \\
\end{gathered} \right)} } \right)n^{\,j} } \hfill \\
\end{gathered}
$$
Bringing the convolution a few steps forward, we can realize that the pattern
of the upper and lower terms of the binomials follows the same pattern as that of
$$
\text{conv}\left( {\left( \begin{gathered} n \\ l \\
\end{gathered} \right) - \left( \begin{gathered} 0 \\ l \\
\end{gathered} \right) - n\left( \begin{gathered} 0 \\ l-1 \\
\end{gathered} \right)} \right) = \sum\limits_{k,\;j} {c_{\,k,\;j} \left( \begin{gathered} k \\ l-j \\
\end{gathered} \right)} \quad \overset {\text{pattern}} \longleftrightarrow \quad \left[ {x^{\;k} y^{\;j} } \right]\text{mult}\left( {x^{\,n} - 1 - ny} \right)
$$
so that we arrive to:
$$
\begin{gathered}
C_{\,2} (m,n,l)\quad \left| {\;0 \leqslant \text{integer}\;m,n,l} \right.\quad = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\;j\,\,\left( { \leqslant \,m} \right)} {\left( { - 1} \right)^{\,j} \left( \begin{gathered}
m \\ j \\
\end{gathered} \right)\left( {\sum\limits_{\left( {0\, \leqslant } \right)\;k\,\,\left( { \leqslant \,m} \right)} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
m - j \\ k \\
\end{gathered} \right)\left( \begin{gathered}
\left( {m - j - k} \right)n \\ l-j \\
\end{gathered} \right)} } \right)n^{\,j} } = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\;j\,\,\left( { \leqslant \,m} \right)} {\left( { - 1} \right)^{\,m - j} \left( \begin{gathered}
m \\ j \\
\end{gathered} \right)\left( {\sum\limits_{\left( {0\, \leqslant } \right)\;k\,\,\left( { \leqslant \,m} \right)} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
j \\ k\\
\end{gathered} \right)\left( \begin{gathered}
\left( { j - k} \right)n \\ l-m+j \\
\end{gathered} \right)} } \right)n^{\,m - j} } \hfill \\
\end{gathered} \tag {5}
$$
which I could check on computer for a small range of the parameters and looks correct.
This time, for $m=0$, by definition the result should be null, for whichever values of $n$ and $l$.
The formula above instead gives $C_{2}(0,n,0) = 1$, which does not look plausible.
So we shall limit its validity to $1 \leqslant m$, or correct it for $m=l=0$.
For an algebraic verification of 5), let us consider an additional relation.
We can partition the $C(m,n,l)$ arrangements with no empty box into those that contain
- $0$ boxes with exactly 1 ball, ${m \choose 0}$ ways to choose them, and $C_{2}(m,n,l)$ boxes with at least $2$ balls
- $1$ boxes with exactly 1 ball, ${m \choose 1}$ ways to choose the box and $n$ ways to place the ball in it, and $C_{2}(m-1,n,l-1)$ boxes with at least $2$ balls
...
- $k$ boxes with exactly 1 ball, ${m \choose k}$ ways to choose the box and $n^k$ ways to place the balls , and $C_{2}(m-k,n,l-k)$ boxes with at least $2$ balls
...
- $m$ boxes with exactly 1 ball, ${m \choose m}$ ways to choose the box and $n^m$ ways to place the balls , and $C_{2}(0,n,l-m)$ boxes with at least $2$ balls.
We can see here that , for $l=m$, the $n^m$ arrangements with exactly $1$ ball per box are legitimate and that this demands that $$C_{2}(0,n,0) = 1$$
The above translates into:
$$
C(m,n,l) = \sum\limits_{\left( {0\, \leqslant } \right)\;k\,\,\left( { \leqslant \,m} \right)} {\left( \begin{gathered}
m \\
k \\
\end{gathered} \right)n^{\,k} \,C_{\,2} (m - k,n,l - k)} \tag {6}
$$
Computing this relation numerically, for $0 \leqslant n,m \leqslant 10$ and $0 \leqslant l \leqslant nm$ returns a full check, and it is also fully demonstrable algebraically.
Then each addendum in formula 6) is exactly what you are looking for.
