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?

Burnside's Lemma states that the number of orbits $|X/G|$ of a set $X$ under the action of a group $G$ is given by:
\begin{equation}
|X/G| = \frac{1}{|G|}\sum_{g \in G}|X^g| 
\end{equation}
where $X^g$ denotes the set of elements in $X$ fixed under the action of $g$.
For the example given, your set $X$ is all the possible ways to arrange three different coloured plates around a square table and $G$ is the set of rotations of the square.
You can think of an arrangement as a string of length four, e.g. the string $RRWB$ indicates the first plate is red, the plate opposite is white, etc. There are $3^4$ arrangements of this kind, so we expect our answer to be smaller than this as, for instance, 
$RRWB,RWBR,WBRR \text{ and } BRRW$ are all considered to be equivalent as each may be obtained from another by rotation.
We say that the set $\{RRWB, RWBR, WBRR, BRRW\}$ is the orbit of the element $RRWB \in X$ under $G$, and the orbit of $X$ is the set of orbits of each element $x \in X$.
Now, $G$ contains $4$ elements which are rotation by $0,90,180 \text{ and } 270$ degrees respectively. Clearly, all elements in $X$ are fixed under rotation by $0$ rads, so $|X^0|=|X|=3^4$.
The arrangements fixed under rotation by $90$ degrees require that each plate be the same colour as the one next to it and thus include $RRRR,WWWW$ and $BBBB$ only. By symmetry, $$|X^{90}|=|X^{270}|=3$$ 
A similar counting argument shows that $|X^{180}|=3^2$ and thus by Burnside's Lemma:
$$|X/G|=\frac{1}{4}\left( |X^0|+|X^{90}|+|X^{180}|+|X^{270}|\right) $$
$$~~ = \frac{1}{4}(81+3+9+3)=\frac{96}{4}=24$$
