combinatorial argument and by induction proof Let n be a fixed natural number. Show that:
$$\sum_{r=0}^m \binom {n+r-1}r = \binom {n+m}{m}$$
(A): using a combinatorial argument and (B): by induction on $m$?

For the sake of completeness I add option (C): using generating functions.
We have
$$\sum_{r=0}^m \binom{n+r-1}{r} = 
\sum_{r=0}^m [z^n] \frac{z}{(1-z)^{r+1}} =
[z^{n-1}] \sum_{r=0}^m \left(\frac{1}{1-z}\right)^{r+1} \\ =
[z^{n-1}] \frac{1}{1-z} \frac{1-1/(1-z)^{m+1}}{1-1/(1-z)} =
[z^{n-1}] \frac{1-1/(1-z)^{m+1}}{1-z-1} =
[z^{n-1}] \frac{1}{z} \left(\frac{1}{(1-z)^{m+1}}-1\right) \\ =
[z^n]  \left(\frac{1}{(1-z)^{m+1}}-1\right).$$
Now there are two cases: first, for $m=0$, we obtain
$$ [z^n]  \left(\frac{1}{(1-z)^1}-1\right) =1 = \binom{n+m}{m}.$$
For $m\ge 1$, and keeping in mind that $n$ is a natural number, we get
$$[z^n]  \left(\frac{1}{(1-z)^{m+1}}-1\right) =
[z^{n+1}] \frac{z}{(1-z)^{m+1}} = \binom{n+1+m-1}{m} =  \binom{n+m}{m}.  $$
