Distinguishable groups of balls into distinguishable boxes with constraints We have $E$ balls of $N$ colors. Let's call $e_i$ the number of balls of color $i$ (of course across group balls are distinguishable while within a color they are not). We can split these balls among $N$ labelled (i.e. distinguishable) jars.
I would like to count the number of ways the balls can be distributed in the jars such as:


*

*the $i$-th jar contains exactly $o_i$ balls

*the $j$-th color is divided exactly among $k_j$ jars


Both conditions should hold. For simplicity we can initially assume $e_i = o_i = k_i$, but the dependency on $i$ should remain.
I know that the problem is highly not-trivial so even a sketch (or a non closed form) solution is wellcome! 
 A: Hint:
Seems to understand that, called $n_{\,i,\,j}$ the number of balls of color $j$
into box $k$, we have a $N \times N$ matrix in which the sums of
the columns, as well as those of the rows are determined as $e_j$ and $o_i$.
Moreover, the $e_j$ shall be equally parted into $k_j$ boxes.
We can summarize the whole as 
$$
\left\{ \matrix{
  1 \le i,j \le N \hfill \cr 
  u_{\,i}  \in \left\{ {0,1} \right\} \hfill \cr 
  \left. \matrix{
  k_{\,j} \backslash e_{\,j} ,\quad m_{\,j} \backslash e_{\,j} ,\quad e_{\,j}  = m_{\,j} \,k_{\,j}  \hfill \cr 
  n_{\,i,\,j}  = u_{\,i} \,m_{\,j}  \hfill \cr}  \right\}\quad  \Rightarrow \quad n_{\,i,\,j}  \in \left\{ 0 \right\} \cup \left\{ {m_{\,j} :m_{\,j} \backslash e_{\,j} } \right\} \hfill \cr 
  \sum\limits_i {n_{\,i,\,j} }  = e_{\,j} ,\quad \sum\limits_j {n_{\,i,\,j} }  = o_{\,i} ,\quad \sum\limits_{i,\,j} {n_{\,i,\,j} }  = E \hfill \cr}  \right.
$$
where of course all the variables are non-negative integers.
Now, it is not clear what you want to keep fixed: only $N$ and $E$ ? or else ?
2nd step
In trying and find, at least, a "possible" approach to the problem, I am presenting a second step.
So, as you commented you are going to keep $e_j$ and $o_i$ as fixed.
Then we may write
$$
\left\{ \matrix{
  1 \le i,j \le N \hfill \cr 
  e_{\,j}  = m_{\,j} \,k_{\,j}  \hfill \cr 
  u_{\,i,j}  \in \left\{ {0,1} \right\},\;\;n_{\,i,\,j}  = u_{\,i,\,j} \,m_{\,j}  \hfill \cr 
  e_{\,j}  = \sum\limits_i {n_{\,i,\,j} }  = m_{\,j} \sum\limits_i {u_{\,i,\,j} } \quad  \Rightarrow \quad \sum\limits_i {u_{\,i,\,j} }  = k_{\,j}  \hfill \cr 
  o_{\,i}  = \sum\limits_j {n_{\,i,\,j} }
  = \sum\limits_j {u_{\,i,\,j} \,m_{\,j} } \quad  \Rightarrow \quad o_{\,i}  \le \sum\limits_j {\,m_{\,j} }  \hfill \cr}  \right.
$$
and with an obvious matrix symbolism we can put it as
$$
\left\{ \matrix{
  \overline {\bf u} \;{\bf U} = \overline {\bf k}  \hfill \cr 
  {\bf U}\;{\bf m} = {\bf o} \hfill \cr 
  \overline {\bf u} \;{\bf U}\;{\bf m} = \overline {\bf k} \;{\bf m} = \overline {\bf u} \;{\bf o} = E \hfill \cr}  \right.
$$
3rd step
The previous can be considered a linear system in the unknowns $u_{i,j}$ (which can be only $0,1$), and given the  $o_i$'s
and the $m_j$'s and $k_j$'s, which are related between them as to give $m_jk_j=e_j$. 
Starting with $N=2$, and then passing to $N=3$, the system above can be rewritten as

where a possible recursive pattern is appearing.
