On the nonnegative integer solutions that satisfy $\sum x_i \leq n$ 
We have the following nice problem: Search for all vectors
$(x_1,\ldots,x_r)$ with $x_i > 0$ that satisfy
$$ x_1+x_2+x_3+\cdots+x_r \leq n $$

$\textbf{discussion}$
One approach I have in mind is as follows: First, consider solutions to $\sum^r x_i = n$ and we know there are ${n-1 \choose r-1}$ such solutions by stars and bars argument. but, before we continue, for this argument to work we must have $n \geq r$. Now, we count how many vectors satisfy $\sum x_i = n-1$ and we know again there are ${n-2 \choose r-1}$ such vectors. If we keep this process down to $r$, that is, if we try to find solutions to $\sum x_i = r $, then we obtain just one $(1,1,1,\ldots,1)$ or ${r-1 \choose r-1}$. Therefore, in total we have
$$ {n-1 \choose r-1 } + {n-2 \choose r-1} + {n-3 \choose r-1} + \cdots + {r-1 \choose r-1} $$
such vectors. Is there a way to simplify this expression? Is there any other approach to tackle this problem?
 A: Suppose that you want to count the subsets of $[n]=\{1,\ldots,n\}$ of size $r$. One way to count them is to divide them up according to the largest number in the subset. If $k$ is the largest number in the subset, there are $\binom{k-1}{r-1}$ ways to choose the $r-1$ smaller numbers in the subset, so there are $\binom{k-1}{r-1}$ subsets of $[n]$ of size $r$ whose largest element is $k$. The possible choices for $k$ range from $r$ through $n$, so $[n]$ has
$$\sum_{k=r}^n\binom{k-1}{r-1}$$
subsets of size $r$. Of course we know that it has $\binom{n}r$ subsets of size $r$, so
$$\sum_{k=r}^n\binom{k-1}{r-1}=\binom{n}r\,;$$
that’s your closed form. This is the hockey stick identity; it comes up a lot and is worth knowing.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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By definition, the answer is given by
\begin{align}
&\bbox[10px,#ffd]{\sum_{s = r}^{n}\braces{%
\sum_{x_{1} = 1}^{\infty}\ldots\sum_{x_{r} = 1}^{\infty}
\bracks{x_{1} + \cdots + x_{r} = s}}}
\\[5mm] = &\
\sum_{s = 0}^{n - r}\sum_{x_{1} = 0}^{\infty}\ldots\sum_{x_{r} = 0}^{\infty}
\bracks{z^{s}}z^{x_{1} + \cdots + x_{r}} =
\sum_{s = 0}^{n - r}\bracks{z^{s}}\pars{\sum_{x = 0}^{\infty}z^{x}}^{r} =
\sum_{s = 0}^{n - r}\bracks{z^{s}}\pars{1 \over 1 - z}^{r}
\\[5mm] = &\
\bracks{z^{0}}\pars{1 - z}^{-r}\sum_{s = 0}^{n - r}
\pars{1 \over z}^{s} =
\bracks{z^{0}}\pars{1 - z}^{-r}\,{\pars{1/z}^{n - r + 1} - 1 \over 1/z - 1}
\\[5mm] = &\
\bracks{z^{0}}\pars{1 - z}^{-r - 1}\pars{z^{r - n} - z} =
\bracks{z^{n - r}}\pars{1 - z}^{-r - 1} =
{-r - 1 \choose n - r}\pars{-1}^{n - r}
\\[5mm] = &\
{n \choose n - r} =
\bbox[10px,#ffd,border:1px groove navy]{n \choose r} \\ &
\end{align}
A: I think you have a good idea in stars and bars, but a couple of modifications make it simpler.  Add another bin for overflow, so that we when we distribute $n$ objects into the bins, there will be at most $n$ in the first $r$ bins.  Also, since we want each of the first $r$ bins to have at least one object, put an object into each of the first $r$ bins to start with, then distribute the remaining $n-r$ objects into $r+1$ bins.  Stars and bars gives the answer $$\binom  nr$$
