Combinatorics proof using bijection. 
This is the 'equation' I need to prove. I am trying to represent the LHS with a Venn Diagram, but I am having troubles with it. 
I defined the them as A, B, and C respectively and stated that B is a subset of A and C is A-B. However, where would 'i' go? i is not part of (at least to me it seems like) any of the sets. 
Thanks in advance:)
 A: \begin{eqnarray*}
\binom{n}{r} \binom{r}{r-i} \binom{n-r}{s-i} &=& \frac{n!}{{r!} {(n-r)!}} \frac{{r!}}{(r-i)!i!} \frac{ {(n-r)!}}{(s-i)!(n-r-s+i)!} 
\end{eqnarray*}
\begin{eqnarray*}
&=& \frac{n!}{\color{red}{(r+s-2i)!} \color{blue}{(n-r-s+2i)!}} \frac{\color{blue}{(n-r-s+2i)!}}{i!(n-r-s+i)!} \frac{ \color{red}{(r+s-2i)!}}{(r-i)!(s-i)!} \\
\end{eqnarray*}
\begin{eqnarray*}
&=& \binom{n}{r+s-2i} \binom{n-r-s+2i}{i} \binom{r+s-2i}{r-i}. 
\end{eqnarray*}
A: You have to be careful doing this. I am going to do it left to right, so each step is considering one binomial as an operation of getting subsets. Consider the following:
LHS:
First you divide $[n]$ into $(A,B)$ such that $|A|=r,|B|=n-r,$ in the second step you divide $A$ into $C,D$ such that $|C|=i$ so you have sets $(B,C,D)$ with cardinals $(n-r,i,r-i)$ then you select $s-i$ from set $B$ you get sets $E,F$ with cardinals $|E|=s-i,|F|=n-r-s+i$ so at the end of this side you get sets $(C,D,E,F)$ cardinals $(i,r-i,s-i,n-r-s+i).$
RHS:
First take $(A,B)$ with cardinals $((r-i)+(s-i),n-((r-i)+(s-i))),$ in the second step take $B$ and divide into $C,D$ with cardinals $(i,n-((r-i)+(s-i))-i)$ so you have $(A,C,D)$ with cardinals $((r-i)+(s-i),i,n-((r-i)+(s-i))-i)$ in the third step take $A$ and divide into $E,F$ with sizes $r-i$ and $s-i$ so you get $(C,D,E,F)$ with sizes $(r-i,s-i,i,n-((r-i)+(s-i))-i).$
So you produce in the two ways the same amount of sets with the same cardinalities. So the numbers should be the same.
A: To make the Venn Diagram, first think about the combinations involved in this question. Often such problems can be solved by interpreting the question as being about choosing "committees", and then computing in two different ways.
Suppose we have $n$ politicians. We want to form a government of $r$ of them. Of these $r$, we want $r-i$ of them to form a "cabinet". Of the $n-r$ politicians outside the government, we want $s-i$ of them to form a "shadow cabinet" (i.e. an "opposition"). We compute the number of ways to do this in two different ways.
First, we choose $r$ of the $n$ politicians to form the government. Of these $r$, choose $r-i$ to form the cabinet. Of the $n-r$ politicians outside the government, choose $s-i$ of them to form the "shadow cabinet". Multiplying these three choices together gives the LHS $$\binom{n}{r} \binom{r}{r-i} \binom{n-r}{s-i}$$
Now let's do it a different way. First we select all those politicians who will be part of some cabinet, either the shadow or governmental one. Thus, we are choosing $(r-i)+(s-i)=r+s-2i$ politicians from the $n$. Now from this collection of $r+s-2i$, we choose $r-i$ to form the cabinet (the remaining $s-i$ make the shadow cabinet). Finally we need to complete the government. Recall that the government consists of $r$ politicians, with $r-i$ of them in the cabinet. So from the $n-r-s+2i$ politicians which are not in cabinets, we choose $i$ to complete the government. Multiplying these three choices together gives the RHS
$$\binom{n}{r+s-2i} \binom{r+s-2i}{r-i} \binom{n-r-s+2i}{i}$$
