Let $[x^p]~f(x)$ mean the co-efficient of $x^{p}$ in $f(x)$.
$$S=\sum_{i=0}^{n} i {m-i \choose p} =[x^p] \sum_{i=0}^{n} i (1+x)^{m-i}$$
The above series is AGP, using its summation, we get
$$S=[x^p]~(1+x)^{m-n}\left(\frac{-1-x+nx+(1+x)^n+x(1+x)^n}{x^2}\right)$$
$$\implies [x^{p+2}]~(1+x)^{m-n}(-1-x+nx+(1+x)^n+x(1+x)^n)$$
$$\implies S=-{m-n \choose p+2}+(n-1) {m-n \choose p+1}+{m \choose p+2}+{m \choose p+1}$$
$$S=-{m-n+1 \choose p+2}+n{m-n \choose p+1}+{m+1 \choose p+2}$$
For the given problem $m=99, n=69, p=30$, then the first twp coefficirnt disappear and we get
$$S={100 \choose 32}.$$
Finally, $q=32$.
Let $\omega=\exp\big(i\frac{\pi}3\big)\cdot$
Observe that, for any integer $k$ we have $\sum_{n=0}^5 \omega^{nk}=
\begin{cases}
\frac{\omega^{6k-1}}{\omega^k-1}=0&\text{if}\ k\notin6\mathbb{Z},\\[3pt]
6&\text{if}\ k\in6\mathbb{Z}.
\end{cases}$
From the Newton binomial formula
\begin{eqnarray*}
\sum_{n=0}^5(1+\omega^n)^{32}
&=&\sum_{n=0}^5\sum_{k=0}^{32}\binom{32}{k}\omega^{nk}
=\sum_{k=0}^{32}\binom{32}{k}\sum_{n=0}^5\omega^{nk}\\
&=&\kern-0.7em\sum\limits_{k\in\{0,6,12,18,24,30\}}\kern-0.6em \binom{32}{k}\times6\\
&=&6\sum_{r=0}^5\binom{32}{6r}.
\end{eqnarray*}
So the required sum is $S=\big(\sum_{n=0}^5(1+\omega^n)^{32}\big)\big/6$.
Now we have $1+\omega^s
=\left\{
\begin{array}{rcl}
2&\text{if}\ s=0,\\[3pt]
\sqrt3\exp\big(i\frac\pi6\big)&\text{if}\ s=1,\\[3pt]
\exp\big(i\frac\pi3\big)&\text{if}\ s=2,\\[3pt]
0&\text{if}\ s=3,\\[3pt]
\exp\big({-}i\frac\pi3\big)&\text{if}\ s=4,\\[3pt]
\sqrt3\exp\big({-}i\frac\pi6\big)&\text{if}\ s=5.
\end{array}
\right.$
In the end we obtain
\begin{eqnarray*}
S&{}={}&\bigg[2^{32}
+3^{16}\!\exp\Big({-}i\frac{2\pi}3\Big)
+\exp\Big(i\frac{2\pi}3\Big)
+0
+\exp\Big({-}i\frac{2\pi}3\Big)
+3^{16}\!\exp\Big(i\frac{2\pi}3\Big)\bigg]\mathbin{\Big/}6\\
&{}={}&\frac{2^{32}-3^{16}-1}6\\
&{}={}&708653429.
\end{eqnarray*}
Edit (to answer a comment below).
I don't see any valuable generalization of the method above.
Suppose that one replaces 32 by $m$, 6 by $\lambda$ and puts $\ell=\lfloor m\lambda\rfloor$, in order to calculate $\sum_{n=0}^\ell\binom{m}{\lambda n}$.
We would take $\omega=\exp(2i\pi/\lambda)$, but the complication is that there's no reason in general to have $\sum\limits_{n\leqslant\ell}\omega^{kn}=0$ for $k\notin\lambda\mathbb{Z}$, as above.
In general this sum would be $\frac{\omega^{k(\ell+1)}-1}{\omega^k-1}\cdot$
With the values $(m,\lambda)=(32,6)$ given in the problem, everything went fine because $\ell+1\in\lambda\mathbb{Z}$ (i. e. $5+1\in6\mathbb{Z}$.
Perhaps the general $\displaystyle\sum_{n=0}^\ell\binom{m}{\lambda n}$ can be calculated by other means?
Best Answer
Start from $$ \frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=0}^{n}(-1)^r \binom{n+3}{r+3} $$Reindex: $$ =\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=3}^{n+3}(-1)^{r-3} \binom{n+3}{r} $$ $$ =\frac{-3!}{(n+1)(n+2)(n+3)}\sum_{r=3}^{n+3}(-1)^{r} \binom{n+3}{r} $$Now add and subtract the terms $0\leq r\leq 2$: $$ =\frac{-3!}{(n+1)(n+2)(n+3)}\left(\sum_{r=0}^{n+3}(-1)^{r} \binom{n+3}{r}-\sum_{r=0}^{2}(-1)^{r} \binom{n+3}{r}\right) $$The binomial theorem graciously takes care of the first series for us, and then the result falls out: $$ =\frac{3!}{(n+1)(n+2)(n+3)}\left(0+\sum_{r=0}^{2}(-1)^{r} \binom{n+3}{r}\right) $$ $$ =\frac{3!}{(n+1)(n+2)(n+3)}\cdot \frac{(n+1)(n+2)}{2} = \frac{3}{n+3} $$