Concerning a "closed" ( = finite summation) formula, start from
$$
\eqalign{
& N_b (s - n,m - 1,n) = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
{\rm 1} \le {\rm integer}\;x_{\,j} \le m \hfill \cr
x_{\,1} + x_{\,2} + \; \cdots \; + x_{\,n} = s \hfill \cr} \right. = \cr
& = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
0 \le {\rm integer}\;y_{\,j} \le m - 1 \hfill \cr
y_{\,1} + y_{\,2} + \; \cdots \; + y_{\,n} = s - n \hfill \cr} \right. \cr}
$$
where $N_b$ is given by
$$
N_b (s,r,m)\quad \left| {\;0 \leqslant \text{integers }s,m,r} \right.\quad =
\sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s}{r+1}\, \leqslant \,m} \right)}
{\left( { - 1} \right)^k \binom{m}{k}
\binom
{ s + m - 1 - k\left( {r + 1} \right) }
{ s - k\left( {r + 1} \right)}\ }
$$
as explained in this related post.
Then the number of ways to obtain a sum $x \le s$ is given by
$$
\eqalign{
& M(x,m,n) = \sum\limits_{x\, \le \,s\,\left( { \le \,m\,n} \right)\,} {N_b (s - n,m - 1,n)} = \cr
& = m^{\,n} - \sum\limits_{0\, \le \,s\, \le \,x - 1\,} {N_b (s - n,m - 1,n)} = \cr
& = m^{\,n} - \sum\limits_{0\, \le \,s\, \le \,x - n - 1\,} {N_b (s,m - 1,n)} = \cr
& = m^{\,n} - \sum\limits_{0\, \le \,s\, \le \,x - n - 1\,} {\sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{s \over m}\, \le \,n} \right)}
{\left( { - 1} \right)^k \left( \matrix{ n \hfill \cr
k \hfill \cr} \right)\left( \matrix{
s + n - 1 - k\,m \cr
s - k\,m \cr} \right)} } = \cr
& = m^{\,n} - \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{s \over m}\, \le \,n} \right)} {\left( { - 1} \right)^k \left( \matrix{
n \hfill \cr
k \hfill \cr} \right)\left( \matrix{
x - 1 - k\,m \cr
x - n - 1 - k\,m \cr} \right)} \cr}
$$
and in fact $M(90,20,5)=3003$.
Note that, as explained in the cited related post, the problem has the geometric equivalent of finding:
the number of integral points on the diagonal plane $y_1, \cdots, y_n=s-n$, intercepted by a $n$-D cube
with side $[0,m-1]$.
The formula for $N_b$ corresponds to calculating the points on the whole plane ($k=0$) and subtracting those
pertaining to the surrounding cubes.
The geometric analogy clearly shows that $N_b(nm-s,m,n)=N_b(s,m,n)$.