Show Burnside's Lemma (Weighted)? Let $G$ be a group of permutations of $X$, and let $I(x)$ be an expression which is constant on each orbit of $G$, so that $$I(g(x))=I(x) \text{ for all } g \in G, x \in X.$$
Let $D$ be a set of representative, one from each orbit, and let $E=\{(g,x) | g(x)=x\}$.
By evaluating the sum $$\sum\limits_{(g,x)\in E} I(x)$$ in two different ways, show that
$$\sum\limits_{x\in D} I(x)= \frac{1}{|G|} \sum\limits_{g\in G} \sum\limits_{x\in F(g)} I(x).$$
This question is a bit overwhelming to me. What exactly would a representative of an orbit be? How would I begin to tackle this problem? And could anyone hint me the two different ways to solve it? 
Edit: Forgot to mention that $F(g)$ is the set of objects fixed by $g$.

It is a double sum, over $g$ and over $x$, so like any double sum it can be evaluated in two different ways.
If you make $x$ the inner variable in the sum, then you get exactly the right hand side of the equality you are trying to prove, but without the $1/|G|$ factor. (Note that $x \in F(g)$ by definition of the set $E$.)
So you have to do the sum taking $g$ as the inner variable, and try and get the left hand side of the equality times $|G|$. For a fixed $x$, there are $|{\rm Stab}_G(x)|$ elements $g$ with $(g,x) \in E$, so this works out as $$\sum_{x \in X}|{\rm Stab}_G(x)|I(x).$$
See if you can take it from there, using the Orbit-Stabilizer Theorem.
