Some curious binomial coefficient identities I was playing around with some polynomials involving binomial coefficients, and inadvertently proved the following two identities:
(1) For all $p, q \in \mathbb{N}$ and $i \in \left[ 0, \min\{p, q\} \right]$:
$$
\begin{pmatrix}p \\ i\end{pmatrix} = \sum_{j=0}^i (-1)^j \begin{pmatrix}q \\ j\end{pmatrix} \begin{pmatrix}p + q - j \\ i - j\end{pmatrix} \text{.}
$$
(2) For all $q \in \mathbb{N}_{\ge 1}$ and $i \in [0, q]$:
$$
\frac{q - i}{q} = \sum_{j=0}^i (-1)^j \begin{pmatrix}i \\ j\end{pmatrix} \begin{pmatrix}2q - 1 - j \\ i - j\end{pmatrix} \begin{pmatrix}q - j \\ i - j\end{pmatrix}^{-1}  \text{.}
$$
Can either of these identities be proven in any trivial way (e.g., by reduction to known identities)?
 A: The first one is just inclusion-exclusion in the following way:
Take the set $[p+q]=\{1,\cdots ,p+q\},$ so you want to take $i$ elements from those such that they all belong to $[p].$ By definition you just restrict yourself to the set $[p]$ and hence there are $\binom{p}{i}$, but on the other hand it is the same as this $$|T\setminus \bigcup _{j=1}^q A_{p+j}|,$$
where $T$ is take all possible subsets of size $i$ from $[p+q]$ which can be done in $\binom{p+q}{i}$ and $A_r = \{S\subset [p+q]:|S|=i \wedge r\in S\}$ (so we are taking out all sets that contain elements on $[p+q]\setminus [p]$.)
By inclusion-exclusion then $$|T\setminus \bigcup _{j=1}^q A_{p+j}|=|T|-\sum _{j=1}^q(-1)^{j-1}\sum _{X\in \binom{[p+q]\setminus [p]}{j}}|\bigcap _{y\in X} A_{y}|,$$
as seeing before, $|T|=\binom{p+q}{i},$ and $|A_r|=\binom{p+q-1}{i-1}$ and if you take $r_1,r_2\in [p+q]\setminus [p],$ $|A_{r_1}\cap A_{r_2}|=\binom{p+q-2}{i-2}$ because you have already chosen $2$, hence the intersections are homogeneous and then $$|T|-\sum _{j=1}^q(-1)^{j-1}\sum _{X\in \binom{[p+q]\setminus [p]}{j}}|\bigcap _{y\in X} A_{y}|=\binom{p+q}{i}-\sum _{j=1}^q(-1)^{j-1}\sum _{X\in \binom{[p+q]\setminus [p]}{j}}\binom{p+q-j}{i-j}=\binom{p+q}{i}-\sum _{j=1}^q(-1)^{j-1}\binom{q}{j}\binom{p+q-j}{i-j},$$
what is you identity.
The second one seems more challenging(but it suggests a probabilistic approach.)
Added: The second is a particular case of the first one. 
Notice that $\frac{\binom{i}{j}}{\binom{q-j}{i-j}}=\frac{(q-i)!i!}{j!(q-j)!}=\frac{\binom{q}{j}}{\binom{q}{i}},$ so on the LHS you get $$\binom{q-1}{i},$$
and then take $p=q-1$ in your first identity and the result follows.
A: For what it's worth in verifying that
$${p\choose r} = \sum_{j=0}^r (-1)^j 
{q\choose j} {p+q-j\choose r-j}$$
we may introduce
$${p+q-j\choose r-j} = {p+q-j\choose p+q-r} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{r-j+1}}
\frac{1}{(1-z)^{p+q+1-r}}
\; dz$$
which vanishes when $j\gt r$ so  we may extend $j$ to infinity and get
for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{r+1}}
\frac{1}{(1-z)^{p+q+1-r}}
\sum_{j\ge 0} (-1)^j {q\choose j} z^j
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{r+1}}
\frac{1}{(1-z)^{p+q+1-r}}
(1-z)^q
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{r+1}}
\frac{1}{(1-z)^{p+1-r}}
\; dz$$
This evaluates by inspection to
$${r+p-r\choose p-r} = {p\choose p-r} = {p\choose r}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\ds{\sum_{j = 0}^{i}\pars{-1}^{\,j}{q \choose j}{p + q - j \choose i - j} =
{p \choose i}.\qquad p, q \in \mathbb{N}\,;\quad i \in \bracks{0,\min\braces{p,q}}}$.

From the above conditions it's clear, because $\ds{j \leq i}$, that
$\ds{p + q - j \geq 0}$. Then, $\ds{{p + q - j \choose i - j}_{j\ >\ i} = 0}$ such that $\color{#f00}{the\ above\ sum\ can\ be\ extended}$ to all numbers $\ds{\in \mathbb{N}}$:
\begin{align}
&\sum_{j = 0}^{i}\pars{-1}^{\,j}{q \choose j}{p + q - j \choose i - j} =
\sum_{j = 0}^{\color{#f00}{\infty}}\pars{-1}^{\,j}{q \choose j}\
\overbrace{{-p - q + i- 1 \choose i - j}\pars{-1}^{i - j}}^{\ds{p + q - j \choose i - j}}
\\[5mm] = &\
\pars{-1}^{i}\sum_{j = 0}^{\infty}{q \choose j}
\braces{\vphantom{\huge A}\bracks{z^{i - j}}\pars{1 + z}^{-p - q + i - 1}} =
\pars{-1}^{i}\bracks{z^{i}}\braces{\vphantom{\huge A}%
\pars{1 + z}^{-p - q + i - 1}\
\overbrace{\sum_{j = 0}^{\infty}{q \choose j}z^{j}}^{\ds{\pars{1 + z}^{q}}}}
\\[5mm] = &\
\pars{-1}^{i}\bracks{z^{i}}\pars{1 + z}^{-p + i - 1} =
\pars{-1}^{i}{-p + i - 1 \choose i}
\\[5mm] = &\
\pars{-1}^{i}\
\underbrace{{-\bracks{-p + i - 1} + i - 1 \choose i}\pars{-1}^{i}}
_{\ds{-p + i - 1 \choose i}}\ =\
\bbx{\ds{p \choose i}}
\end{align}
