Proof by induction for ${m+n \choose n+r} = \sum\limits_{k=0}^{m}{m \choose k}{n \choose r+k}$

Question

I've gotten as far as splitting it into two sets and I got: $${m+n+1 \choose m+r} = \sum\limits_{k=0}^{m}{m \choose k}\left({n+1 \choose r+k}\right) = \sum\limits_{k=0}^{m}{m \choose k}\left({n \choose r+k}+{n\choose r+k-1}\right)= {m+n\choose m+r}+\sum\limits_{k=0}^{m}{m\choose k}{n\choose r+k-1} $$ but I'm stuck about how to solve the final term. I also tried doing induction on m but got stuck in a similar place.

(I do not think that Vandermonde's identity does helps with this problem, as I am looking to solve for ${m+n \choose m+r}$ rather than ${m+n \choose r}$.)

Just to clarify, is there some difference between the meaning of $\left({n+1 \choose r+k}\right)$ and $\binom{n+1}{r+k}$? Similarly $\left(\binom n{r+k}\right)$ and $\binom n{r+k}$ in your post on meta. — Martin Sleziak, Oct 10 '19 at 03:53
It seems that using Approach0 you can find a few posts that are very close to yours. For example: Prove that $ \sum_{k=0}^m {m \choose k} {n \choose r+k} = {m+n \choose m+r}$, Prove that $\sum\limits_{k=0}^m\binom{m}k\binom{n}{r+k}=\binom{m+n}{m+r}$ using combinatoric arguments., ... — Martin Sleziak, Oct 10 '19 at 03:58
... Prove by Double Counting Method $\sum\limits_{k = 0}^m \binom{m}{k}\binom{n}{r + k} = \binom{m + n}{m + r}$, Show that $\sum_{k=0}^{\infty} \binom{r}{k} \binom{s}{n+k} = \binom{r+s}{r+n}$. — Martin Sleziak, Oct 10 '19 at 03:59
@JMoravitz Since the question was closed by you, I thought it might be reasonable to let you know about the OP's post on meta: I don't believe my question is a duplicate — Martin Sleziak, Oct 10 '19 at 04:05
Since I have mentioned Approach0, I will also add link to a post with some other tips on searching: How to search on this site? — Martin Sleziak, Oct 10 '19 at 05:47

Proof by induction for ${m+n \choose n+r} = \sum\limits_{k=0}^{m}{m \choose k}{n \choose r+k}$

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