# Another counting problem on the number of ways to place $l$ balls in $m$ boxes.

As it seems that this was a long question, I broke it down into two simpler ones which you may see here and here.

I need some help to afford the following counting problem. I appreciate any help in advance.

There are $m$ boxes. Each box is partitioned to $n$ parts. There are also $l$ balls where $\ m\ <\ l\ <\ (m-1)n+2$.

I would like to know how many ways there are to place the $l$ balls in the $m$ boxes such that

$1-$ Each part of a box can hold up to one ball.

$2-$ Each box must have at least one ball.

I could handle this part, and I found the answer as $$C(m,n,l)\ =\ \sum_{i=0}^{k}(-1)^i \binom{m}{i}\binom{(m-i)n}{l},$$ where $k = m - \lceil l/n\rceil$.

Now, I need to know the number of cases that exactly $j$ boxes contain only one ball, and the other boxes contain more than one ball. Fore sure, I mean for the possible amounts for $j$.
I initially guessed that the answer is $$\binom mj C(m-j, n, l-m).$$
Actually, I selected $j$ boxes and put one ball in each one. So, I have now $l-m$ balls which should be placed on the other $m-j$ boxes. To do it, I used the obtained function above ($C(m,n,l)$). However, I think there is something wrong with my answer.