Number of distinct solutions of $x_1+x_2+\dots+x_k=n$ with $0\le x_i\le r$ Let $n,k,r$ be positive integers. Let $S(n,k,r)$ be the number of all  solutions of the Diophantine Equation $x_1+x_2+\dots+x_k=n$ with $0\le x_i\le r$ for every $i\in \{1,2,\dots,k\}$?
How many of these solutions are distinct, that is, $x_i\ne x_j$ for $i\ne j$?

Let $T(n,k,r)$ be this number. Let $s\le k$. There are $\binom{k}{s}$ ways to select $s$ unknowns $x_i$. The number of solutions in which these $s$ unknowns are equal, seems to be
$$\sum_{i=0}^{\lfloor \frac{n}{s}\rfloor}T(n-si,k-s,r)$$
So it seems:
$$T(n,k,r)=S(n,k,r)-\sum_{s=2}^n\sum_{i=0}^{\lfloor \frac{n}{s}\rfloor}\binom{k}{s}T(n-si,k-s,r).$$
Can we derive a formula for $T(n,k,r)$ from it?
It doesn't seem to be a good approach. Assuming $S(n,k,r)$ is known, is there a formula for $T(n,k,r)$?

For future reference here is  a formula using the cycle index $Z(P_k)$
of the unlabeled set operator $\mathfrak{P}_{=k}:$
$$[z^n] Z(P_k)\left(1+z+\cdots +z^r\right).$$
This cycle index has OGF
$$G(w) = 
\exp\left(a_1z-a_2\frac{z^2}{2}+a_3\frac{z^3}{3}-\cdots\right)
= \exp\left(\sum_{l\ge 1} (-1)^{l-1} a_l \frac{w^l}{l}\right).$$
Doing the substitution for this particular case we obtain
$$H(z, w) = \exp\left(\sum_{l\ge 1} (-1)^{l-1} \frac{w^l}{l}
\sum_{q=0}^r z^{lq}\right)
\\ = \exp\left(\sum_{q=0}^r 
\sum_{l\ge 1} (-1)^{l-1} \frac{w^l}{l} z^{lq}\right)
\\ = \exp\left(\sum_{q=0}^r \log(1+wz^q)\right)
\\ = \prod_{q=0}^r (1+wz^q).$$
We get for the answer
$$[w^k] [z^n] H(z, w)
= [w^k] [z^n] \prod_{q=0}^r (1+wz^q)$$
which matches what was observed in the comments.
Multiply by $k!$ if permutations are considered distinct.
Remark. Obviously this does not need PET and follows straightforwardly by inspection.
