Li Shanlan's combinatorial identities I am struggling to prove the following combinatorial identities:
$$(1)\quad\sum_{r=0}^m \binom{m}{r}\binom{n}{r}\binom{p+r}{m+n} = \binom{p}{m}\binom{p}{n},\quad \forall n\in\mathbb N,p\ge m,n$$
$$(2)\quad\sum_{r=0}^m \binom{m}{r}\binom{n}{r}\binom{p+m+n-r}{m+n} = \binom{p+m}{m}\binom{p+n}{n},\quad \forall p\in\mathbb N$$
The book I found them in says they were discovered and proved by Chinese mathematician Li Shanlan in the 19th century but I have failed to find any of his works translated into English in the Internet.
I am looking for an either combinatorial or algebraic solution.
 A: Solution for $(1)$:
$$\begin{align}
\sum_{r=0}^m \binom{m}{r}\binom{n}{r}\color{blue}{\binom{p+r}{m+n}}
&=\sum_{r=0}^m \binom{m}{r}\binom{n}{r}\color{blue}{\sum_{j}\binom{p}{m+j}\binom r{n-j}}\\
&= \sum_{r=0}^m\sum_{j} \binom{m}{r}\binom{p}{m+j}\color{orange}{\binom{n}{r}\binom r{n-j}}\\
&= \sum_{r=0}^m\sum_{j} \binom{m}{r}\binom{p}{m+j}\color{orange}{\binom{n}{n-j}\binom j{r-n+j}}\\
&= \sum_{r=0}^m\sum_{j} \binom{m}{r}\binom{p}{m+j}\color{orange}{\binom nj\binom j{n-r}}\\
&= \sum_{j} \binom{p}{m+j}\binom nj\color{purple}{\sum_{r=0}^m\binom{m}{r}\binom j{n-r}}\\
&= \sum_{j} \binom{p}{m+j}\binom nj\color{purple}{\binom{m+j}n}\\
&= \sum_{j} \color{green}{\binom{p}{m+j}\binom{m+j}n}\binom nj\\
&= \color{green}{\binom pn}\sum_{j} \color{green}{\binom{p-n}{p-m-j}}\binom nj\\
&= \binom pn\binom{p}{p-m}\\
&=\binom pn \binom pm=\binom pm \binom pn\qquad\blacksquare
\end{align}$$

Solution for $(2)$:
Put   $\; p=1-q\;$ in $(1)$ above:
$$\begin{align}
\sum_{r=0}^m\binom mr \binom nr\binom {1-q+r}{m+n}
&=\binom {1-q}m\binom {1-q}n
\\
\sum_{r=0}^m\binom mr \binom nr (-1)^{m+n}\binom {m+n+q-r}{m+n}
&=(-1)^m\binom {m+q}m (-1)^n\binom {m+q}n
\qquad\text{using upper negation}
\\
\sum_{r=0}^m \binom mr\binom nr \binom {m+n+q-r}{m+n}
&=\binom {m+q}m \binom{n+q}n
\\
\sum_{r=0}^m \binom mr\binom nr \binom {m+n+p-r}{m+n}
&=\binom {p+m}m\binom {p+n}n
\quad \text{putting $p$ for $q$ WLOG}\qquad\blacksquare\end{align}$$
A: Very striking identities. Hard to guess how Li Shanlan discovered them...
Concerning (1), multiply by $x^n y^m z^p$ and add up. 
In the rhs we have
$$
\sum_{p\ge 0}z^p\sum_{n=0}^p {p \choose n} x^n \sum_{m=0}^p {p \choose m} y^m  = \sum_{p\ge 0}(1+x)^p(1+y)^p z^p = \frac{1}{1-z(1+x)(1+y)}.
$$
The lhs is tougher. Denoting $t = \frac{z}{1-z}$,
$$
\sum_{r\ge 0}\sum_{n\ge r}{n\choose r}x^n \sum_{m\ge r}{m\choose r}y^m \sum_{p\ge m+n-r} {p+r\choose m+n}z^p \\
= \sum_{r\ge 0}\sum_{n\ge r}{n\choose r}x^n \sum_{m\ge r}{m\choose r}y^m\frac{z^{m+n-r}}{(1-z)^{m+n+1}} = \sum_{r\ge 0}\frac1{z^r(1-z)}\sum_{n\ge r}{n\choose r}(tx)^n \sum_{m\ge r}{m\choose r}(ty)^m\\
= \sum_{r\ge 0}\frac1{z^r(1-z)}\frac{(tx)^r}{(1-tx)^{r+1}}\frac{(ty)^r}{(1-ty)^{r+1}} = \frac{1}{(1-z)(1-tx)(1-ty)}\frac{1}{1-\frac{1}{z}\frac{tx}{1-tx}\frac{ty}{1-ty}}\\
= \frac{1}{(1-tx)(1-ty)\left(1-z-\frac{txy}{(1-tx)(1-ty)}\right)} = \frac{1}{(1-z)(1-tx)(1-ty) - txy}\\
= \frac{1}{(1-z-xz)(1-ty) - txy} = \frac{1}{1-xz -yz - txy + txyz} \\
= \frac{1}{1-z-xz -yz - xyz} = \frac{1}{1-z(1+x)(1+y)}.
$$
A: Assume $n\geq m$. $\binom{p}{n}\binom{p}{m}$ is the coefficient of the monomial $x^n y^m$ in the expansion of $(1+x)^p(1+y)^p=(1+x+y+xy)^p$, i.e. the coefficient of $x^n y^m$ in:
$$ \sum_{k=0}^{p}\binom{p}{k}(x+y+xy)^k=\sum_{k=0}^{p}\binom{p}{k}\sum_{j=0}^{k}\binom{k}{j}x^j y^j (x+y)^{k-j}$$
and we may restrict the last sum over $m+n=k+j$, having:
$$ \sum_{j=0}^{m}\sum_{\substack{k\geq j \\ k+j=m+n}}\binom{k}{j}\binom{p}{k}\binom{k-j}{n-j}=\sum_{j=0}^{m}\binom{m+n-j}{j}\binom{p}{m+n-j}\binom{m+n-2j}{n-j} $$
that can be rearranged in the wanted form $(1)$.
A: Suppose we seek to verify that
$$\sum_{r=0}^{\min\{m,n,p\}} 
{m\choose r} {n\choose r}
{p+m+n-r\choose m+n}
= {p+m\choose m} {p+n\choose n}.$$
Introduce
$${n\choose r} = {n\choose n-r} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-r+1}} 
(1+z)^n \; dz$$
and
$${p+m+n-r\choose m+n} = {p+m+n-r\choose p-r} \\ =
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{p-r+1}} 
(1+w)^{p+m+n-r} \; dw.$$
Observe carefully  that the first of  these is zero when  $r\gt n$ and
the second  one when $r\gt  p$ so  we may extend  the range of  $r$ to
infinity.

This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n+1}} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p+m+n}}{w^{p+1}} 
\sum_{r\ge 0} {m\choose r} z^r \frac{w^r}{(1+w)^r}
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n+1}} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p+m+n}}{w^{p+1}} 
\left(1+\frac{zw}{1+w}\right)^m
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n+1}} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p+n}}{w^{p+1}} 
(1+w+zw)^m
\; dw \; dz.$$
The inner integral is
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p+n}}{w^{p+1}} 
\sum_{q=0}^m {m\choose q} (1+z)^q w^q
\; dw$$
with residue
$$\sum_{q=0}^{\min(m,p)} {m\choose q} {p+n\choose p-q}
(1+z)^q$$
which in combination with the outer integral yields
$$\sum_{q=0}^{\min(m,p)} {m\choose q} {p+n\choose n+q}
{n+q\choose n}.$$
Now note that
$${p+n\choose n+q} {n+q\choose n}
= \frac{(p+n)!}{(p-q)! (n+q)!} \frac{(n+q)!}{q! n!}
\\ = \frac{(p+n)!}{(p-q)! p!} \frac{p!}{q! n!}
= {p+n\choose n} {p\choose q}.$$
Therefore we just need to verify that
$$\sum_{q=0}^{\min(m,p)} {m\choose q} {p\choose p-q}
= {p+m\choose m}$$
which follows by inspection.
It can also be done with the integral
$${p\choose p-q} =
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p}}{w^{p-q+1}} \; dw$$ 
which is zero when $q\gt p$ so we can extend $q$ to infinity
to get for the sum
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p}}{w^{p+1}} 
\sum_{q\ge 0} {m\choose q} w^q
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p+m}}{w^{p+1}} 
\; dw
\\ = {p+m\choose m}.$$ 
