Is there another way to represent this summation? I wish to calculate $\sum_{x=1}^{n}\sum_{y=1}^{n} f(x,y)$ where $x>2y$. I can do this by changing $y$'s upperbound to the floor of $(x-1)/2$ but this makes simplification of the summation harder later. Is there a way using inclusion-exclusion to simplify this sum?
 A: No need to evaluate a floor.  Since $x > 2y$, $x \geq 2y + 1:$
$$\sum_{y=1}^n \sum_{x=2y+1}^n f(x,y).$$
A: According to my own experience, for the purpose of simplifying a double sum, the first and most tricky and powerful step is to suitably extend the defining domain of $f(x,y)$ so that we have 
$$\sum_{x=1}^n\sum_{y=1}^nf(x,y)
=\sum_{x=-\infty}^{+\infty}\sum_{y=-\infty}^{+\infty}f(x,y),$$
which is usually written in the notation
$$\sum_{x}\sum_{y}f(x,y),$$
for the sake of emphasizing that we now have a summation of "standard" form in hand.
In other words, we simply define $f(x,y)=0$ when $x<1$ or $x>n$ or $y<1$ or $y>n$. For this problem, we can define $f(x,y)=0$ for $x\le 2y$. The remaining work is routine, just to simply
$\sum_{x}\sum_{y}f(x,y)$
by using standard transformations and known formulas. For example, instead of writing 
$$\sum_{k=0}^n{n\choose k}=2^n,$$
I prefer the "standard" form
$$\sum_{k}{n\choose k}=2^n.$$
This trick sometimes brings great convenience, and maybe helpful to your specific problem.
