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?

There are four possible rotations of (clockwise) 0, 90, 180 and 270 degrees respectively. Let us denot the 90 degree rotation by $A$, so the other rotations are then its powers $A^i,i=0,1,2,3$. The exponent is only relevant modulo 4, IOW we have a cyclic group of 4 elements. These rotations act on the set of plate arrangements. So if $RWBR$ denotes the arrangement, where there is a Red plate on the North side, White on the East side, Blue on the South, and another Red plate on the Western seat, then
$$
A(RWBR)=RRWB,
$$
because rotating the table 90 degrees clockwise moves the North seat to East et cetera.
The idea in using Burnside's lemma is to calculate how many arrangements are fixed under the various rotations (or whichever motions you won't count as resulting in a distinct arrangement). Let's roll.
All $3^4=81$ arrangements are fixed under not doing anything to the table. So $A^0$ has $81$ fixed points.
If an arrangement stays the same upon having a 90 degree rotation act on it, then the plate colors at North and East, East and South, South and West, West and North must all match. IOW we use a single color only. Therefore the rotation $A^1$ has only $3$ fixed points: RRRR, WWWW and BBBB.
The same applies to the 270 degree rotation $A^3$. Only 3 fixed points.
The 180 degree rotation $A^2$ is more interesting. This rotation swaps the North/South and East/West pairs of plates. For an arrangement to be fixed under this rotation, it is necessary and sufficient that those pairs of plates had matching colors, but we can use any of the three colors for both N/S and E/W, so altogether there are 9 arrangements stable under the 180 degree rotation.
The Burnside formula then tells that the number of distinguishable table arrangements is
$$
\frac14(81+3+9+3)=\frac{96}4=24.
$$
