How to prove that $\sum_{i=0}^{a}\frac{i\binom{a+b-c-i}{a-i}\binom{c+i-1}{i}}{\binom{a+b-1}{a}}=\frac{ac(a+b)}{b(b+1)}$ let $$b\ge c,a,b,c\in N^{+}$$
Show that
$$\sum_{i=0}^{a}\dfrac{i\binom{a+b-c-i}{a-i}\binom{c+i-1}{i}}{\binom{a+b-1}{a}}=\dfrac{ac(a+b)}{b(b+1)}$$
This sum is similar to Hypergeometric distribution,
but different.
I know this
$$\sum_{i=0}^{a}\binom{a}{i}\binom{b}{c-i}=\binom{a+b}{c}$$
so
$$\sum_{i=0}^{a}\binom{a+b-c-i}{a-i}\binom{c+i-1}{i}=\sum_{i=0}^{a}\binom{a+b-c-i}{b-c}\binom{c+i-1}{c-1}=\binom{a+b-1}{a}？$$
so
$$\sum_{i=0}^{a}\dfrac{\binom{a+b-c-i}{a-i}\binom{c+i-1}{i}}{\binom{a+b-1}{a}}=1$$
use this
$$k\binom{n}{k}=n\binom{n-1}{k-1}$$
$$i\binom{c+i-1}{i}\binom{a+b-c-i}{a-i}=(c+i-1)\binom{c+i-2}{i-1}\binom{a+b-c-i}{a-i}$$
Then I can't .Thank you 
 A: Your result follows from some standard results about binomial coefficients
(as found in [1], for example).
I'm not sure of the minimal conditions needed to make the argument work,
 but I will assume that $a\geq 0$ and  $b>c\geq 1$. I first rewrite your sum, without
the factor $i$ as
$$\sum_{i\geq 0}{a+b-c-i\choose b-c}{c-1+i\choose c-1}={a+b\choose b}.\tag1$$
This follows from an application of (5.26) from [1].
We will now exploit the fact that the sum in (1) is valid for all $a\geq 0$. 
Let's  re-introduce the factor of $i$ and define 
$$f(a)=\sum_{i\geq 0}{a+b-c-i\choose b-c} i {c-1+i\choose c-1}.$$
Then
\begin{eqnarray*}
f(a)
&=& \sum_{i\geq 1}{a+b-c-i\choose b-c} i {c-1+i\choose i}\\[5pt]
&=& \sum_{i\geq 1}{a+b-c-i\choose b-c} [c+(i-1)] {c+i-2\choose i-1}\\[5pt]
&=& \sum_{j\geq 0}{a-1+b-c-j\choose b-c} [c+j] {c+j-1\choose j}\\[5pt]
&=& c{a-1+b\choose b}+ f(a-1). 
\end{eqnarray*}
Since $f(0)=0$, we deduce that 
$$f(a)=c\sum_{k=0}^{a-1}{k+b\choose b}=c{a+b\choose a-1},$$
where the last equation comes from page 174 of Concrete Mathematics
under "parallel summation". 
It is not hard to check that 
$$c{a+b\choose a-1}={\binom{a+b-1}{a}}\dfrac{ac(a+b)}{b(b+1)}$$
so this gives the desired result. 
Reference [1]  Concrete Mathematics (2e) by  Graham, Knuth, and Patashnik. 
