Possible ways to have $n$ bounded natural numbers with a fixed sum Is it possible to count in an easy way the solutions of the equations and inequalities $x_1+x_2+\cdots+x_n = S$ and $x_i\leq c_i$ if all $x_i$ and $c_i$ are natural numbers?

We know that the number of non-negative integral solutions to the system:
$\begin{cases} x_1+x_2+x_3+\dots+x_n = S\\
0\leq x_1\\
0\leq x_2\\
\vdots\\
0\leq x_n\end{cases}$
is $\binom{S+n-1}{n-1}$, seen by stars-and-bars counting.  In the event that you do not consider zero to be a natural number, then it is as though you have a lower bound of $1\leq x_i$ for each so make a change of variable as $y_i=x_i-1$ to get it into the above form.
Denote all solutions where upper bounds are ignored as $\Omega$ and designate this as our universal set.  With upper bounds on each, continue via inclusion-exclusion.  Let $A_i\subset \Omega$ be the event that $x_i>c_i$, I.e. $x_i\geq c_i+1$.  The set of solutions that you are interested in then is $(\bigcup A_i)^c$.  The total being $|(\bigcup A_i)^c| = |\Omega|-|(\bigcup A_i)|$
This can be broken apart further as you wish, but becomes tedious to write in a compact formula.  Counting $|A_1\cap A_2\cap A_3|$ for example will be solutions to the system:
$\begin{cases} x_1+x_2+\dots+x_n = S\\
c_1+1\leq x_1\\
c_2+1\leq x_2\\
c_3+1\leq x_3\\
0\leq x_4\\
\vdots\\
0\leq x_n\end{cases}~~~~~~\Leftrightarrow \begin{cases} y_1+y_2+\dots+y_n = S-c_1-c_2-c_3-3\\
0\leq y_1\\
0\leq y_2\\
\vdots\\
0\leq y_n\end{cases}$
using the change of variable $y_i = x_i-c_i-1$ for $i\in\{1,2,3\}$ and $y_i = x_i$ otherwise.
An attempt to write in a compact form, using indexing set notation for intersections and sums, and letting $N=\{1,2,3,\dots,n\}$
$$\sum\limits_{\Delta\subseteq N}\left((-1)^{|\Delta|}|\bigcap_{i\in\Delta}A_i|\right) = \sum\limits_{\Delta\subseteq N}\left((-1)^{|\Delta|}\binom{S+n-1-|\Delta|-\sum\limits_{i\in\Delta}c_i}{n-1}\right)$$
of course, using that $\binom{n}{r} = 0$ whenever $n<r$ or $n<0$ or $r<0$, i.e. the original form of the binomial coefficient, not the generalized form.
