number of solutions to $x_1+x_2+...+x_m=n$ 
Question:
What is the number of solutions to the equation:
$x_1+x_2+\cdots+x_m=n$
when $\forall \ i : 1\le x_i\le n$

My approach:
I took $\ $ $\forall \ i : y_i= x_i-1 \ $ and then i got the equation:
$\ y_1+y_2+\cdots+y_m=n-m \ $ when $\ \forall \ i : 0\le y_i\le n-1$.
And now I take $\ $ $\forall \ i : z_i= n-1-y_i \ $ so now the equation will be:
$\ z_1+z_2+\cdots+z_m=n-1-y_1+\cdots+n-1-y_m=m(n-1)-(y_1+\cdots+y_m)=m(n-1)-(n-m)=mn-m-n+m=n(m-1) \ $
to make a long story short:
$\ z_1+\cdots+z_m=n(m-1) \ $ when $\ \forall \ i : 0\le z_i$
and for such an equation the number of solutions is simply $\binom{(n+1)(m-1)}{n(m-1)}$
please help me. Did I make some mistakes or is it a good solution.
 A: Your solution is incorrect.  
Method 1:  Solving the problem in the non-negative integers.  
We wish to find the number of solutions to the equation
$$x_1 + x_2 + \cdots + x_m  = n \tag{1}$$
where each $x_k$, $1 \leq k \leq m$, is a positive integer.  As you observed, if we set $y_k = x_k - 1$, then $y_k$ is a non-negative integer and 
$$y_1 + y_2 + \cdots + y_m = n - m \tag{2}$$
which has the same number of solutions in the non-negative integers that equation 1 has in the positive integers.  A particular solution of equation 2 corresponds to the placement of $m - 1$ addition signs in a row of $n - m$ ones.
For example, if $n = 10$ and $m = 3$, then $n - m = 7$, so 
$$1 1 + + 1 1 1 1 1$$
corresponds to the solution $y_1 = 2$, $y_2 = 0$, and $y_3 = 5$, while
$$1 1 1 1 + 1 1 + 1$$ 
corresponds to the solution $y_1 = 4$, $y_2 = 2$, and $y_3 = 1$.  
Thus, the number of solutions of equation 2 is the number of ways we can select which $m - 1$ of the $(n - m) + (m - 1) = n - 1$ symbols ($n - m$ ones and $m - 1$ addition signs) are addition signs, which is 
$$\binom{n - 1}{m - 1}$$
Method 2:  Solving the problem in the positive integers.
A particular solution of equation 1 in the positive integers corresponds to the placement of $m - 1$ addition signs in the $n - 1$ spaces between successive ones in a row of $n$ ones.  Hence, the number of solutions of equation in the positive integers is 
$$\binom{n - 1}{m - 1}$$
For example, if $n = 10$ and $m = 3$.  Then the solution 
$$1 1 1 + 1 + 1 1 1 1 1 1$$
corresponds to $x_1 = 3$, $x_2 = 1$, and $x_3 = 6$, while the solution
$$1 1 1 1 1 + 1 1 1 + 1 1$$
corresponds to $x_1 = 5$, $x_2 = 3$, and $x_3 = 2$. Notice that these cases correspond to the two examples given above when we solved the problem in the non-negative integers.   
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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With $\ds{\quad\pars{~x_{i} \in \mathbb{N}_{\ \geq\ 1}\,,\quad i = 1, 2, \ldots, m~},\quad}$ the number of configurations is given by:

\begin{align}
&\sum_{x_{1}\ =\ 1}^{\infty}\cdots\sum_{x_{m}\ =\ 1}^{\infty}
\bracks{x_{1} + \cdots + x_{m} = n} =
\sum_{x_{1}\ =\ 1}^{\infty}\cdots\sum_{x_{m}\ =\ 1}^{\infty}\,\,\,
\oint_{\verts{z}\ =\ 1^{-}}\,\,\,{1 \over z^{n + 1 - x_{1} - \cdots - x_{m}}}\,
\,\,\,\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\oint_{\verts{z}\ =\ 1^{-}}\,\,\,{1 \over z^{n + 1}}
\pars{\sum_{x = 1}^{\infty}z^{x}}^{m}\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z}\ =\ 1^{-}}\,\,\,{1 \over z^{n + 1}}\,\pars{z \over 1 - z}^{m}
\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\oint_{\verts{z}\ =\ 1^{-}}\,\,\,{1 \over z^{n - m + 1}}\,\,\pars{1 - z}^{-m}
\,\,\,{\dd z \over 2\pi\ic} =
\bracks{z^{n - m}}\pars{1 - z}^{-m}
\\[5mm] = &\
{-m \choose n - m}\pars{-1}^{n - m} =
\braces{{m +\bracks{n - m} - 1\choose n - m}\pars{-1}^{\, n - m}}
\pars{-1}^{\, n - m} = \bbx{\ds{n - 1 \choose n - m}}
\end{align}
