How to prove this using Combinatorial proof?

Question

Here is the problem:

$\displaystyle\sum_{k=0}^{p}\binom{p+q-k}{q}\binom{r+k}{r}\ = \binom{p+q+r+1}{p}$

How to prove this using combinatorics? I don't want to use algebra or something. My idea was that since we are not explicitly choosing $k$, we probably shouldn't choose from a subset which can be added to $r$ or subtracted from $p+q$,instead we can imagine a bit string of probably p+q+r+1 length with $r+1$ ones and $p+q$ zeroes and consider counting on basis of occurrence of last one. But it isn't working out to be correct. Can someone please help

Answer 1

Imagine choosing $q+r+1$ numbers from $1,...,p+q+r+1$ such that the $r+1$ - th smallest number is $r+k+1$.

1. Choose $r$ numbers from $1,...,r+k$, $\binom{r+k}{r}$
2. $r+k+1$ is already determined as the $r+1$ - th number
3. Then choose remaining $q$ numbers from $r+k+2,...,p+q+r+1$, $\binom{p+q-k}{q}$

In total, there are $\binom{p+q-k}{q}\binom{r+k}{r}$ possibilities. Now what if we sum for all possible values of $k$? We get the number of all possible $q+r+1$ combination from $1,...,p+q+r+1$. Isn’t that $\binom{p+q+r+1}{q+r+1}=\binom{p+q+r+1}{p}$?