What is the number of equivalence classes of $\{0,\ldots,n\}^m/\mathrm S_m$? 
What is the number of equivalence classes of $\{0,\ldots,n\}^m/\mathrm S_m$?

where $\mathrm S_m$ is the symmetric group and the equivalence relation is (obviously?) defined as
$$\alpha\sim\beta\iff \alpha\in\beta\cdot\mathrm S_m$$
Im trying to calculate it but it seems hard. Suppose that $n+1\ge m$.
Then if Im not wrong for some subset $A\subset\{0,\ldots,n\}$ such that $|A|=m$ then if $\alpha\in\{0,\ldots,m\}^m$ then
$$|A^m/\mathrm S_m|=\sum_{|\alpha|=m}\binom{m}{\alpha}=m^m$$
where $|\alpha|:=\sum \alpha_j$, in other words the above is the sum of all multinomial coefficients for $m$ positions and $m$ possibly different elements.
If $n+1>m$ then exist $\binom{n+1}{m}$ subsets of cardinality $m$, then the final answer should be
$$|\{0,\ldots,n\}^m/\mathrm S_m|=\binom{n+1}{m}\sum_{|\alpha|=m} \binom m \alpha =\binom{n+1}{m}m^m$$
But if $n+1<m$ then the multinomial sum will be not complete and we can calculate it in a more complicated way and where is not a closed form.
Im very unsure of my calculations. Can someone confirm or correct it? Thank you in advance.

Thank to both answers, both are correct, at least to this point:
$$|\{0,\ldots,n\}^m/\mathrm S_m|=\sum_{k=1}^{n+1}\binom{n+1}{k}\binom{m-1}{k-1}=\binom{m+n}{m}$$


*

*The Vandermonde's identity is key to show the equality.

 A: Elements of $\{0,\ldots,n\}^m$ consist of $m$-tuples and I am assuming $S_m$ acts by place permutation. Since we are only interested in the orbits under the action of $S_m$, we can choose a nice representative:
$$(\underbrace{0,\ldots,0}_{m_0},\underbrace{1,\ldots,1}_{m_1},\ldots,\underbrace{n,\ldots,n}_{m_n}).$$
Note that size of the orbit of this element is $$\displaystyle{m\choose m_0,\ldots,m_n}=\frac{m!}{m_0!m_1!\cdots m_n!}$$
and the number of such orbits is equal to the number of compositions of $m$ into at most $n+1$ parts, with parts labelled by an increasing sequence of integers between $0$ and $n$. Alternatively, one can count the number of orbits as the number of partitions of $m$ with at most $n+1$ parts, with parts labelled by distinct integers (in any order) between $0$ and $n$. The number of partitions of $m$ with exactly $k$ parts is the coefficient of $x^m$ in the power series expansion of
$$\frac{x^k}{(1-x)(1-x^2)\cdots(1-x^{k})}.$$
It follows that the number of labelled partitions of $m$ with exactly $k$ parts is the coefficient of $x^m$ in power series expansion of
$$\frac{{n+1\choose k}x^k}{(1-x)(1-x^2)\cdots(1-x^{k})}$$
and, therefore, the number of orbits is the coefficient of $x^m$ in the power series expansion of
$$\sum_{k=0}^{n+1}\frac{{n+1\choose k}x^k}{(1-x)(1-x^2)\cdots(1-x^{k})}.$$
A: To  keep it  simple  we count  the number  of  equivalence classes  of
$\{1,2,\ldots,n\}^m$  under the  action of  the symmetric  group $S_m$
acting on the slots. By stars-and-bars we should end up with
$${m+n-1\choose m}.$$
Now by  the Polya  Enumeration Theorem with  $Z(S_m)$ being  the cycle
index of $S_m$ we should get
$$Z(S_m)(A_1+A_2+\cdots+A_n)_{A_1=A_2=\cdots=A_n=1}.$$
This is
$$\frac{1}{m!} \sum_{p=1}^m \left[m\atop p\right] n^p.$$
Evaluating this with the OGF of the Stirling numbers of the first kind
which is
$$\sum_{k=1}^m \left[m\atop k\right] z^k
= \prod_{q=0}^{m-1} (z+q)$$
we find
$$\frac{1}{m!} \sum_{p=1}^m n^p 
[z^p] \left(m! \times {z+m-1\choose m}\right)
= {n+m-1\choose m}.$$
Remark. Since the  issue with this question appears  to be getting
the interpretation right and ascertaining what exactly is being asked,
I am  posting some relevant Maple  code. Consult the code  to discover
what interpretation  I am  using. I  have deliberately  refrained from
simplification  and optimization  in order  to represent  the original
problem as stated.

S :=
proc(n,m)
option remember;
local ind, orbits, dg, items;

    orbits := table();

    for ind from n^m to 2*n^m-1 do
        dg := convert(ind, base, n);

        items := [seq(dg[q], q=1..m)];
        orbits[convert(items, `multiset`)] := 1;
    od;

    nops([indices(orbits)]);
end;

T := (n,m)-> binomial(m+n-1,m);

