Prove that the number of pairs $(A,B)$ equals ${{n}\choose{i}}{{n-i}\choose{r-i}}{{n-r}\choose{s-i}}$ Prove that the number of pairs $(A,B)$ with $A\subseteq N_n, B\subseteq N_n, |A|=r, |B|=s, and |A\cap B|=i$ equals

$$ {{n}\choose{i}}{{n-i}\choose{r-i}}{{n-r}\choose{s-i}}.$$

My teacher told me that I am supposed to give a proof using combinatorics (so not a purely algebraic one). What I did was draw a Venn-diagram with the sets A and B, and then I think you can interpret ${{n}\choose{i}}$ as the number of combinations from the intersection of A and B, ${{n-i}\choose{r-i}}$ as the combinations from $A - A\cap B$ and ${{n-r}\choose{s-i}}$ as the number of combinations from $B-A\cap B$. However, I am not sure whether this is a complete proof. Could anyone please comment on my approach and/or suggest a different approach and then give the proof? Thank you in advance.
 A: Your approach is good, complete, and fully combinatorial. The Venn diagram is the appropriate guide. 
There are $\binom{n}{i}$ ways to choose who will be in $A\cap B$. For every one of these ways, there are $\binom{n-i}{r-i}$ ways to choose the remaining $r-i$ members of $A$. And once this is done, there are $\binom{n-r}{s-i}$ ways to choose the remaining members of $B$. 
Remark: The expression we get is not optimal in appearance. It can be simplified to the more symmetrical
$$\frac{n!}{(n-r)!(n-s)!(r+s-n)!}.$$
A: Yes, your argument is fine: $\binom{n}i$ is the number of ways of picking which elements of $N_n$ belong to $A\cap B$. Once you know them, you need only pick $A\setminus(A\cap B)$, which can be done in $\binom{n-i}{r-i}$ ways, and the rest of $B\setminus(A\cap B)$, which can then be done in $\binom{n-r}{s-i}$ ways, to determine the sets $A$ and $B$ completely. You might want to expand it a little by pointing out that in the last step you’re choosing $B\setminus(A\cap B)$ from $N_n\setminus A$, since all of $A$ has then been chosen, and that that’s why you’re choosing from a set of $n-r$ elements.
