Each "occupancy histogram" corresponds to a $n$-tuple, which geometrically translates into a point
in the ${\mathbb N}^n$ space, that is, in the non-negative portion of a $n$-dimensional grid.
When the number of balls is fixed (let me indicate it with $s$, to reserve $k$ for use as index),
the points will lay on the diagonal plane $x_1+x_2+ \cdots +x_n=s$.
If the capacity of each bin is unlimited, or however higher than the number of balls, the plane
will extend on all the non-negative portion, and its "area" will be equal to the number
of weak compositions of $s$ into $n$ parts, i.e.
$$
N_b(s,n) = \binom{s+n-1}{s}
$$
If the capacity of each bin is constant and equal to $r$, then the plane will be limited inside
a $n$-cube with sides $[0,r]$, and the intercepted area will be
$$
N_b (s,r,n) = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
{\rm 0} \le {\rm integer}\;x_{\,j} \le r \hfill \cr
x_{\,1} + x_{\,2} + \; \cdots \; + x_{\,n} = s \hfill \cr} \right.
$$
which, as explained in this related post, is given by
$$
N_b (s,r,n)\quad \left| {\;0 \leqslant \text{integers }s,n,r} \right.\quad =
\sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s}{r+1}\, \leqslant \,n} \right)}
{\left( { - 1} \right)^k \binom{n}{k}
\binom
{ s + n - 1 - k\left( {r + 1} \right) }
{ s - k\left( {r + 1} \right)}\ }
$$
If the capacity of each bin is different, then the plane is limited inside a $n$-parallepiped with sides $[0,r_k]$, and
a closed expression for the intercepted area would be in general quite complicated.
Premised the above, we have then to make clear that the process of "pouring" the balls is different than
that of "throwing" the balls into the bin.
While we can understand the "pouring" as the process in which each point ($n$-tuple) is equiprobable,
in "throwing" it is normally understood that we have have $n^s$ equi-probable outcomes (for unlimited capacity)
which is not the same as the $N_B(s,n)$ given above.
That is because the process of throwing, implies a succession and thus a distinction among the balls given
by the order in which they are thrown. You can convince yourself of the difference by taking examples with a small
number of bins and balls.
The throwing process shall be examined through the Stirling development of $n^s$
$$
n^{\,s} = \sum\limits_k {\left\{ \matrix{
s \cr
k \cr} \right\}n^{\,\underline {\,k\,} } } = \sum\limits_k {k!\left\{ \matrix{
s \cr
k \cr} \right\}\left( \matrix{
n \cr
k \cr} \right)}
$$
that is through the number of ways of partitioning a set of $s$ elements into $k$ non empty sub-sets, and distribute
the $k$ subsets into the $n$ bins.
The multinomial approach
$$
\begin{array}{l}
\left( {x_{\,1} + x_{\,2} + \cdots + x_{\,n} } \right)^{\,s}
= \cdots + x_{\,l_{\,1} } x_{\,l_{\,2} } \cdots x_{\,l_{\,s} } + \cdots \quad \left| {\;\,l_{\,j} \in \left[ {1,n} \right]} \right. = \\
= \sum\limits_{\left\{ {\begin{array}{*{20}c}
{0\, \le \,j_{\,l} } \\ {j_{\,1} + \,j_{\,2} + \, \cdots + \,j_{\,n} \, = \,s} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
s \\ j_{\,1} ,\,j_{\,2} ,\, \cdots ,\,j_{\,n} \\
\end{array} \right)\;x_{\,1} ^{j_{\,1} } \;x_{\,2} ^{j_{\,2} } \; \cdots \;x_{\,n} ^{j_{\,n} } } \\
\end{array}
$$
also captures the "throwing" process:
$$
x_{\,l_{\,1} } x_{\,l_{\,2} } \cdots x_{\,l_{\,s} }
$$
stands in fact for 1st ball in bin $l_1$, .., s-th ball in bin $l_s$.
That is more "straight" to deal when setting upper limits to the $j_k$ indices.
In the unlimited case it is equivalent to the Stirling development, since
$$
n!\left\{ \begin{array}{c}
s \\ n \\
\end{array} \right\} = \sum\limits_{\left\{ {\begin{array}{*{20}c}
{1\, \le \,j_{\,l} } \\ {j_{\,1} + \,j_{\,2} + \, \cdots + \,j_{\,n} \, = \,s} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
s \\ j_{\,1} ,\,j_{\,2} ,\, \cdots ,\,j_{\,n} \\
\end{array} \right)\;}
$$
