A sum involving Bernoulli numbers and product of binomial coefficients

Question

I came across a sum involving Bernoulli numbers and product of binomial coefficients during a computation. Precisely, it is $\sum\limits_{c=0}^m B_c\binom{r}{c}\binom{r-c}{m-c}$ where $B_c$ denotes the $c$th Bernoulli number, $r\geq m$, $m\geq 3$, and $m$ odd. Numerically I could evaluate it to be 0 through the sage code given here. Is there anyway to see it in general that this sum is indeed 0 for any general $r$ and $m$ with these properties? I can also see that the binomial product is symmetric from the both the ends of the sum. That is, $\binom{r}{j}\binom{r-j}{m-j}$ and $\binom{r}{m-j}\binom{r-m+j}{j}$ are the same.

Answer 1

We have $\binom{r}{c}\binom{r-c}{m-c}=\frac{r!(r-c)!}{c!(r-c)!(m-c)!(r-m)!}=\frac{r!}{m!(r-m)!}\frac{m!}{c!(m-c)!}=\binom{r}{m}\binom{m}{c}$, then $\sum\limits_{c=0}^m B_c\binom{r}{c}\binom{r-c}{m-c}=\binom{r}{m} \sum\limits_{c=0}^m \binom{m}{c}B_c=0$.

Answer 2

With Mathematica I have only a solution to series:

$$\sum _{c=0}^m B_c \binom{r}{c} \binom{r-c}{m-c}=-\frac{\zeta (1-m) (-1+m-r)!}{(-1+m)! (-1-r)!}$$ for: $m>1$