Suppose we seek to evaluate
$$\sum_{k=0}^n (-1)^k
{1+p+q\choose k} {p+n-k\choose n-k} {q+n-k\choose n-k}$$
which is claimed to be
$${p\choose n}{q\choose n}.$$
Introduce
$${p+n-k\choose n-k}
= \frac{1}{2\pi i}
\int_{|z_1|=\epsilon} \frac{(1+z_1)^{p+n-k}}{z_1^{n-k+1}}
\; dz_1$$
and
$${q+n-k\choose n-k}
= \frac{1}{2\pi i}
\int_{|z_2|=\epsilon} \frac{(1+z_2)^{q+n-k}}{z_2^{n-k+1}}
\; dz_2.$$
Observe that these integrals vanish when $k\gt n$ and we may extend
$k$ to infinity.
We thus obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z_1|=\epsilon}
\frac{(1+z_1)^{p+n}}{z_1^{n+1}}
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon}
\frac{(1+z_2)^{q+n}}{z_2^{n+1}}
\\ \times \sum_{k\ge 0} {1+p+q\choose k} (-1)^k
\frac{z_1^k z_2^k}{(1+z_1)^k (1+z_2)^k}
\; dz_2\; dz_1.$$
This is
$$\frac{1}{2\pi i}
\int_{|z_1|=\epsilon}
\frac{(1+z_1)^{p+n}}{z_1^{n+1}}
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon}
\frac{(1+z_2)^{q+n}}{z_2^{n+1}}
\\ \times
\left(1-\frac{z_1 z_2}{(1+z_1)(1+z_2)}\right)^{p+q+1}
\; dz_2\; dz_1$$
or
$$\frac{1}{2\pi i}
\int_{|z_1|=\epsilon}
\frac{(1+z_1)^{n-q-1}}{z_1^{n+1}}
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon}
\frac{(1+z_2)^{n-p-1}}{z_2^{n+1}}
(1+ z_1 + z_2)^{p+q+1}
\; dz_2\; dz_1$$
Supposing that $p\ge n$ and $q\ge n$ this may be re-written as
$$\frac{1}{2\pi i}
\int_{|z_1|=\epsilon}
\frac{1}{z_1^{n+1} (1+z_1)^{q+1-n}}
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon}
\frac{1}{z_2^{n+1} (1+z_2)^{p+1-n}}
\\ \times (1+ z_1 + z_2)^{p+q+1}
\; dz_2\; dz_1$$
Put $z_2 = (1+z_1) z_3$ so that $dz_2 = (1+z_1) \; dz_3$ to get
$$\frac{1}{2\pi i}
\int_{|z_1|=\epsilon}
\frac{1}{z_1^{n+1} (1+z_1)^{q+1-n}}
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon}
\frac{1}{(1+z_1)^{n+1} z_3^{n+1} (1+(1+z_1)z_3)^{p+1-n}}
\\ \times (1+ z_1)^{p+q+1} (1+z_3)^{p+q+1}
\; (1+z_1) \; dz_3\; dz_1$$
which is
$$\frac{1}{2\pi i}
\int_{|z_1|=\epsilon}
\frac{(1+z_1)^p}{z_1^{n+1}}
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon}
\frac{1}{z_3^{n+1} (1+ z_3 + z_1 z_3)^{p+1-n}}
\\ \times (1+z_3)^{p+q+1}
\; dz_3\; dz_1
\\ = \frac{1}{2\pi i}
\int_{|z_1|=\epsilon}
\frac{(1+z_1)^p}{z_1^{n+1}}
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon}
\frac{(1+z_3)^{n+q}}{z_3^{n+1} (1 + z_1 z_3 /(1+z_3))^{p+1-n}}
\; dz_3\; dz_1$$
Extracting the residue for $z_1$ first we obtain
$$\sum_{k=0}^n {p\choose n-k}
\frac{(1+z_3)^{n+q}}{z_3^{n+1}}
{k+p-n\choose k}
(-1)^k \frac{z_3^k}{(1+z_3)^k}.$$
The residue for $z_3$ then yields
$$\sum_{k=0}^n (-1)^k {p\choose n-k}
{k+p-n\choose k} {n-k+q\choose n-k}.$$
The sum term here is
$$\frac{p!\times (p+k-n)!\times (q+n-k)!}
{(n-k)! (p+k-n)! \times k! (p-n)! \times (n-k)! q!}$$
which simplifies to
$$\frac{p!\times n! \times (q+n-k)!}
{(n-k)! \times n!\times k! (p-n)! \times (n-k)! q!}$$
which is
$${n\choose k} {p\choose n}{q+n-k\choose q}$$
so we have for the sum
$${p\choose n}
\sum_{k=0}^n {n\choose k} (-1)^k {q+n-k\choose q}.$$
To evaluae the remaining sum we introduce
$${q+n-k\choose q}
= \frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{(1+v)^{q+n-k}}{v^{q+1}} \; dv$$
getting for the sum
$${p\choose n}
\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{(1+v)^{q+n}}{v^{q+1}}
\sum_{k=0}^n {n\choose k} (-1)^k \frac{1}{(1+v)^k} \; dv
\\ = {p\choose n}
\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{(1+v)^{q+n}}{v^{q+1}}
\left(1-\frac{1}{1+v}\right)^n \; dv
\\ = {p\choose n}
\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{(1+v)^{q}}{v^{q-n+1}} \; dv
= {p\choose n} {q\choose q-n}$$
which is $${p\choose n} {q\choose n}.$$
This concludes the argument.