Combination with repetitions removing the permutated sequences Is there any closed form expression for the number of combinations with repetitions removing the permutated sequences (for a given number of spaces (N1) and a given number of possible numbers (N2))?
For example in this case there are 5 spaces and 4 possible numbers: 
1 1 1 1 1 -> Valid
1 1 1 1 2 -> Valid
1 2 1 1 1 -> This one is a repetition of the above line
1 2 4 4 1 -> Valid
1 4 1 4 2 -> This one is a repetition of the above line
1 2 3 4 4 -> Valid 

It does not matter getting the sequence 1 2 4 4 1 or de 1 4 1 4 2 in the end. But it is not possible to get both of them.
Is this a known problem? Is there any elegant algorithm or closed form expression to solve this problem?
Thanks!
 A: Suppose there are $r$ spaces, and  $n$ possible numbers $1,...,n$.

Counting only one instance for sequences which permute to each other, each sequence of $r$ terms, where each term is one of the numbers $1,..,n$, corresponds to a solution $(x_1,...,x_n)$ of the equation
$$x_1 + \cdots + x_n = r$$
where each $x_k$ is a nonnegative integer representing the multiplicity of the number $k$ in the given sequence.

By the stars-and-bars formula, the desired count is
$$\binom{r+n-1}{n-1}$$
For example, if $r=5,\;n=4$, we get a count of
$$\binom{5+4-1}{4-1}=\binom{8}{3}=56$$
To illustrate the correspondence, note that the sequence 
$$(1,1,3,4,4)$$
is the unique sequence (up to a permutation of the terms) with 


*

*$2$ terms equal to $1$

*$0$ terms equal to $2$

*$1$ term equal to $3$

*$2$ terms equal to $4$


corresponding to the solution
$$(x_1,x_2,x_3,x_4)=(2,0,1,2)$$
of the equation 
$$x_1+x_2 + x_3 + x_4 = 5$$
A: To present another derivation of the answer already given, you are looking for the
Number of words, of length $r$, with characters taken from the alphabet $\{1,2,\cdots,n \}$,
net of (non considering) the permutations.
Now, not considering the permutations is clearly equivalent to arrange the characters "alphabetically", i.e.
in a non-decreasing (or non-increasing) order.
So you are looking  for
Number of non-decreasing words, of length $r$, with characters taken from the alphabet $\{1,2,\cdots,n \}$.
That is the number of solutions to
$$
1 \le \;x_{\,1}  \le x_{\,2}  \le  \cdots  \le x_{\,r}  \le n
$$
or
$$
0 \le \;y_{\,1}  \le y_{\,2}  \le  \cdots  \le y_{\,r}  \le n - 1
$$
Put
$$
z_{\,1}  = y_{\,1} \quad z_{\,2}  = y_{\,2}  - y_{\,1} \quad  \cdots \quad z_{\,r}  = y_{\,r}  - y_{\,r - 1} 
$$
and you get
$$
0 \le \;z_{\,1}  \le z_{\,1}  + z_{\,2}  \le  \cdots  \le z_{\,1}  + z_{\,2}  +  \cdots  + z_{\,r}  \le n - 1
$$
i.e.
$$
\left\{ \matrix{
  0 \le z_{\,j}  \le n - 1\quad \left| {\;1 \le j \le r} \right. \hfill \cr 
  z_{\,1}  + z_{\,2}  +  \cdots  + z_{\,r}  \le n - 1\quad  \hfill \cr}  \right.
$$
which is the same as
$$
\left\{ \matrix{
  0 \le z_{\,j}  \le n - 1\quad \left| {\;1 \le j \le r + 1} \right. \hfill \cr 
  z_{\,1}  + z_{\,2}  +  \cdots  + z_{\,r}  + z_{\,r + 1}  = n - 1\quad  \hfill \cr}  \right.
$$
and that is the number of weak compositions of $n-1$ into exactly $r+1$ parts.
$$
\left( \matrix{
  n - 1 + r \cr 
  n - 1 \cr}  \right) = \left( \matrix{
  n - 1 + r \cr 
  r \cr}  \right)
$$
