I was solving one problem where I am stuck in the middle... Plz kindly suggest some ways to proceed further.. So the problem is:
The value of
$$\frac{\sum_{i=0}^{100}{{k}\choose{i}}{{m-k}\choose{100-i}}\frac{k-i}{m-100}}{{m}\choose{100}}$$
where :
$m-k > 100$,
$k > 100$ equals ?
- $k$/$m$
- $m$/$k$
- $k$/$m^2$
- $m$/$k$
So to solve this what I did:
I separated the $k-i$ term as $k$ and $-i$ along with other factors and then the term having $k$ was a Vandermondes property but what about the term with $i$??
My problem is how to proceed in those cases where Vandermondes property is multiplied by $i$...?
I wrote the given term as:
$$\frac{\sum{{k}\choose{i}}{{m-k}\choose{100-i}}\frac{k}{m-100}}{{m}\choose{100}}-\frac{\sum{{k}\choose{i}}{{m-k}\choose{100-i}}\frac{i}{m-100}}{{m}\choose{100}} $$
Now here the first term is succinctly Vandermondes formula multiplied with $k$, which is a constant.. So I applied the property and got this:
$$\frac{{{m}\choose{100}}\frac{k}{m-100}}{{m}\choose{100}} -\frac{\sum{{k}\choose{i}}{{m-k}\choose{100-i}}\frac{i}{m-100}}{{m}\choose{100}} $$
Now how to proceed with the second term.. There "i" is multiplied to Vandermondes property ..!! How to simplify in such cases..?? (Plz suggest some good articles / links also for further read on the approaches if possible.)
P.S.
Vandermondes Formula:
$${n+m \choose k} = \sum_{j=0}^{k}{n \choose j}{m \choose k-j}$$
For $i<k$ we have $(k-i)\binom{k}{i}=\frac{k!}{i!(k-i-1)!}=\color{blue}{k}\binom{k-1}{i}$. Thus, denoting $s=100$, your value is $$\binom{m}{s}^{-1}\frac{k}{m-s}\sum_{i=0}^{s}\binom{k-1}{i}\binom{m-k}{s-i}=\binom{m}{s}^{-1}\frac{k}{m-s}\binom{m-1}{s}=\frac{k}{m}.$$