Burnside's Lemma I've been trying to understand what Burnside's Lemma is, and how to apply it, but the wiki page is confusing me. The problem I am trying to solve is:

You have 4 red, 4 white, and 4 blue identical dinner plates. In how
  many different ways can you set a square table with one plate on each
  side if two settings are different only if you cannot rotate the table
  to make the settings match?

Could someone explain how to use it for this problem, and if its not too complicated, try to explain to me what exactly it is doing in general?

The idea behind Burnside's lemma is fairly simple. Given a set $X$ and a group $G$ acting on it, it relates the number of orbits of $X$ under $G$, which are basically the subsets of $X$  which are traced out by $G$, to the number of elements of $X$ fixed by elements of $G$. Rigorously, orbits are sets of the form $\{gx: g\in G\}$ for fixed $x\in X$. The set of elements fixed by $g\in G$, denoted $X^g$ is the set $\{x: x\in X, gx=x\}$. The lemma states $$\text{Number of orbits}=\frac{1}{|G|}\sum\limits_{g\in G} |X^g|$$
which is useful because it's usually easier to count fixed points than it is to count orbits. 
In your case, you are interested in how many ways you can arrange four plates, up to clockwise rotation. This can be stated as the number of sets of the form 
$$\{\text{setting, setting rotated once, setting rotated twice, setting rotated thrice}\}$$
or more compactly $\{abcd, dabc, cdab, bcda\}$ where each letter represents a plate. More mathematically, the number of orbits of the set of arrangements of your plates under the action of $C_4 = \{1,r,r^2,r^3\}$, the group of cyclic permutations. So to apply Burnside's lemma, we need to find the number of arrangements fixed by each element of $C_4$. The identity element fixes all arrangements, so $|X^1|=3^4$ as we can pick one of three colors for each place. The elements fixed by $r$, rotation to the right, are those which are all the same color, of which there are $3$, so $|X^r|=3$. Similarly, $|X^{r^3}|=3$. Finally, elements fixed by $r^2$ are of the form $abab$, so we have two choices of color for a total of $3\times 3=9$ elements. Thus we have 
$$\text{Number of orbits}=\frac{1}{4}\sum\limits_{g\in G}|X^g|=\frac{1}{4}(3^4+3+9+3)=24.$$
