how to solve $\sum _{m=0}^{k-1}mC_{k-1}^{m}C_{N-k}^{m}$? solving
$$\sum _{m=0}^{k-1} mC_{k-1}^m C_{N-k}^m$$
the solution seems to be
$$\frac {\left( N-2\right) !} {\left( k-2\right) !\left( N-k-1\right) !}$$
according to some clue from the other problem.
struggle with this the whole afternoon, please help.
i tried to extend it, but it's too complicated. i think there should be some smart trick to apply on it to make it easier.
i remember i saw this problem long time ago, but i forgot the solution.
UPDATE: the original suspected solution is wrong. now it's corrected.
 A: Suppose we seek to evaluate
$$\sum_{m=0}^{k-1} m {k-1\choose m} {N-k\choose m}.$$
Start from
$${N-k\choose m}
= \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{N-k} \; dz.$$

This gives the following integral for the sum
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon} 
\sum_{m=0}^{k-1} m {k-1\choose m}
\frac{1}{z^{m+1}} (1+z)^{N-k} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^{N-k}}{z}
\sum_{m=0}^{k-1} m {k-1\choose m}
\frac{1}{z^m}  \; dz.$$
Now recall that
$$x ((1+x)^n)' = \sum_{q=0}^n q {n\choose q} x^q$$
and
$$x ((1+x)^n)' = n x (1+x)^{n-1}$$
so that the sum in the integral simplifies to
$$\frac{k-1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^{N-k}}{z}
\frac{1}{z}\left(1 + \frac{1}{z}\right)^{k-2} \; dz.
\\ = \frac{k-1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^{N-k}}{z}
\frac{1}{z}\frac{(1+z)^{k-2}}{z^{k-2}} \; dz
\\ = \frac{k-1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^{N-2}}{z^k} \; dz.$$
The last integral may be evaluated by inspection and gives
$$(k-1) {N-2\choose k-1}
= \frac{(N-2)!}{(k-2)!(N-k-1)!}.$$
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
A: Related techniques: (I). We have

$$ \sum _{m=0}^{k-1} m\,C_{k-1}^m C_{N-k}^m = \frac{1}{(N-k)!}\sum _{m=0}^{k-1} m\,C_{k-1}^m\frac{m!}{(m-(N-k))!}. $$

Let's start with the identity

$$S= \sum_{m=0}^{k-1} C_m^{k-1}x^m = (1+x)^{k-1}\implies \sum_{m=0}^{k-1} m\,C_m^{k-1}x^{m-1} = (k-1)(1+x)^{k-2} $$

Multiplying both sides of the last equation by $x$, differentiating $N-k$ times and dividing by $(N-k)!$ gives
$$ \frac{1}{(N-k)!}\sum_{m=0}^{k-1} mC_m^{k-1}\frac{m!}{(m-(N-k))!}x^{m-(N-k)} = \frac{(k-1)}{(N-k)!}\sum_{i=0}^{N-k} C_{i}^{N-k} x^{(i)}((1+x)^{k-2})^{(N-k-i)}. $$
You can see that on the right hand side of the last equation that all the terms are $0$ except $i=0,1$. So there is nothing left in the problem except to simplify the above and then substitute $x=1$.
Notes: You need the following facts 
1) The $r$th differentiation formula for $x^s$

$$ D^{r}x^{s} = \frac{s!}{(r-s)!} x^{s-r}. $$ 

2) The product formula for differentiation

$$ (fg)^{(r)} \sum_{i=0}^{n}{r \choose i } f^{(i)}g^{(r-i)} $$ 

A: after a few days research, i have come up with a easy arithmetic solution without calculus.
$$\sum _{m=0}^{k-1} mC_{k-1}^m C_{N-k}^m$$
$$=\sum _{m=0}^{k-1} (k-1)C_{k-2}^{m-1} C_{N-k}^m$$ this will be proved by simply extending it.
$$=\sum _{m=0}^{k-1} (k-1)C_{k-2}^{k-m-1} C_{N-k}^m$$
$$=(k-1)C_{N-2}^{k-1}\ ............\ lemma\ 1$$ this will be proved later
$$=\frac{(k-1)(N-2)!}{(k-1)!(N-k-1)!}$$
$$=\frac{(N-2)!}{(k-2)!(N-k-1)!}$$
now i will prove lemma 1.
observe $(1+x)^{a+b}$, and its extension. notice that the factor of term $x^d$ is $C_{a+b}^d$.
then for $(1+x)^{a}$, the factor of term $x^e$ is 
$C_{a}^e$.
then for $(1+x)^{b}$, the factor of term $x^f$ is 
$C_{b}^f$.
let $e+f=d$, list the summation:
$$\sum_{e:\ as\ much\ as\ meaningful}C_{a}^eC_{b}^{d-e}=C_{a+b}^d$$
proved
A: $$Vandermonde's Identity: \sum_{i=0}^{j} \binom{y}{j-i}\binom{x}{i}=\binom{x+y}{j}$$  
$$\sum_{m=0}^{k-1} m\binom{k-1}{m}\binom{N-k}{m}=\sum_{m-1=0}^{k-2} m\binom{k-1}{k-1-m}\frac{N-k}{m}\binom{N-k-1}{m-1}$$
$$\quad\quad\quad\quad=(N-k)\sum_{m-1=0}^{k-2} \binom{k-1}{k-2-(m-1)}\binom{N-k-1}{m-1}$$
$$\quad=(N-k)\binom{N-2}{k-2} (By\;Vandermonde's\; Identity)$$ 
$$\quad=(N-k)\frac{(N-2)!}{(k-2)!(N-k)!}$$
$$\quad=\frac{(N-2)!}{(k-2)!(N-k-1)!}$$
