Another Hockey Stick Identity I know this question has been asked before and has been answered here and here.
I have a slightly different formulation of the Hockey Stick Identity and would like some help with a combinatorial argument to prove it. First I have this statement to prove:
$$
\sum_{i=0}^r\binom{n+i-1}{i}=\binom{n+r}{r}.
$$
I already have an algebraic solution here using the Pascal Identity:
$$
\begin{align*}
\binom{n+r}{r}&=\binom{n+r-1}{r}+\binom{n+r-1}{r-1}\\
&=\binom{n+r-1}{r}+\left[\binom{n+(r-1)-1}{(r-1)}+\binom{n+(r-1)-1}{r-2}\right]\\
&=\binom{n+r-1}{r}+\binom{n+(r-1)-1}{(r-1)}+\left[\binom{n+(r-2)-1}{r-2}+\binom{n+(r-2)-1}{(r-2)-1}\right]\\
&\,\,\,\vdots\\
&=\binom{n+r-1}{r}+\binom{n+(r-1)-1}{(r-1)}+\binom{n+(r-2)-1}{(r-2)-1}+\binom{n+(r-3)-1}{r-3}+\cdots+\left[\binom{n+1-1}{1}+\binom{n+1-1}{0}\right]\\
&=\binom{n+r-1}{r}+\binom{n+(r-1)-1}{(r-1)}+\binom{n+(r-2)-1}{(r-2)-1}+\binom{n+(r-3)-1}{r-3}+\cdots+\binom{n+1-1}{1}+\binom{n-1}{0}\\
&=\sum_{i=0}^r\binom{n+i-1}{i}.
\end{align*}
$$
I have read both combinatorial proofs in the referenced answers above, but I cannot figure out how to alter the combinatorial arguments to suit my formulation of the Hockey Stick Identity. Basically, this formulation gives the "other" hockey stick. Any ideas out there?
 A: Note that $\binom{n+r}{r}=\binom{n+r}{n}$ is the number of subsets of $\{1,2,\ldots,n+r\}$ of size $n$.  On the other hand, for $i=0,1,2,\ldots,r$, $\binom{n+i-1}{i}=\binom{n+i-1}{n-1}$ is the number of subsets of $\{1,2,\ldots,n+r\}$ of size $n$ whose largest element is $n+i$.
A: Suppose that the Diophantine inequality $x_1 + x_2 + ... + x_n \le r$ has $A(n, r)$  non-negative integer solutions. (Or one has to disturb at most $r$ objects into $n$ bins and this task is possible in $A(n, r)$ ways. Note that one distinguishes the bins but one does not wish to distinguish the objects)
We will calculate $A(n, r)$ in two ways.
$$
x_1 + x_2 + ... + x_n \le r 
\\
\Rightarrow \exists x_{n+1} \in \mathbb{Z^+}\cup\{0\}: x_1 + x_2 + ... + x_n + x_{n+1} = r
$$
According to stars and bars problem,
$$
A(n, r) = \left(\!\!{n + 1 \choose r}\!\!\right) =  {n + r \choose r} \qquad \mathcal{\color{navy}{(I)}}
$$
Hence wee seek integer solutions (and $r$ is also an integer), by the rule of sum, $A(n, r)$ would be the sum of non-negative integer solutions to these equations:
$$
x_1 + x_2 + \cdots + x_n = 0\\or\\
x_1 + x_2 + \cdots + x_n = 1\\or\\
x_1 + x_2 + \cdots + x_n = 2\\or\\
\vdots\\or\\
x_1 + x_2 + \cdots + x_n = r
$$
For all $0 \le i \le r$, the equation $x_1 + x_2 + ... + x_n = i$ would have  $\left(\!\!{n \choose i}\!\!\right) =  {n + r - 1 \choose r}$ non-negative integer solutions. Hence,
$$
A(n, r) = \sum_{i=0}^r\left(\!\!{n \choose i}\!\!\right) = \sum_{i=0}^r{n+i-1 \choose i} \qquad \mathcal{\color{navy}{(II)}}
\\
{\color{navy}{(I)}}, {\color{navy}{(II)}} \Rightarrow {n + r \choose r} = \sum_{i=0}^r\left(\!\!{n \choose i}\!\!\right) = \sum_{i=0}^r{n+i-1 \choose i}
$$
