Coloring a rectangle with 3 rows and 4 columns using two colors. I want to color a 3x4 rectangle using 2 colors. The number of squares colored by each color must equal 6. However, we say that two colorings are equal if they can be obtained from each other by permutating rows or by cyclically permutating columns, e.g. $$\matrix{1 & 1 & 0 & 1\\0&1&0&1\\1&0&0&0}=\matrix{0&1&0&1\\0&1&1&1\\0&0&1&0}$$
(we switch the first two rows and then move columns cyclically by two)
The only way I thought of for calculating that was count the total number of such colorings $\binom{12}{6}=924$, and dividing it first by 6 (because each coloring has 6 ways to be permutated), then by 4 (because each coloring has 4 different corresponding cyclic permutations), but that clearly doesn't work, not only because e. g.$$\matrix{0&1&0&0\\1&1&1&1\\0&1&0&0}$$
has only 3 different row permutations, but also because 924 is not divisible by 24. So what is the proper way to solve that (without brute-forcing all 924 colorings)?
 A: Well, one way is to use Burnside's Counting Lemma. There may be easier ways, but I like Burnside's counting lemma.
There are, as you noted, 924 ways to colour the grid if you ignore symmetry. 
The symmetries give a group action on those 924 colorings, you want to count the orbits of that action.
Burnside's counting lemma says that to count the orbits, you can average the number of colourings fixed by each element of the group.
The symmetry group is $S_3\times S_4$, which has 144 elements, so that's a lot of work, but you can simplify things a bit as follows.
Instead of counting orbits of the 924 colourings, you realise that the action of $S_4$ preserves the column counts, so a colouring with 3 in one column, 2 in another, 1 in a third column and 0 in a fourth column will always have those counts, no matter how you permute the rows and columns. So, I'll count colourings with those numbers of coloured squares in the first, second, third and fourth columns respectively, and average the number of such colourings fixed by each element of $S_3$.
There's ${}^3C_3$ ways to colour the first column, ${}^3C_2$ ways to colour the second column, ${}^3C_1$ ways to colour the third column and ${}^3C_0$ ways to colour the last column. That's a total of $1\times3\times3\times1$ colourings. But I need to take symmetries into account.
Well, $S_3$ has six elements. All 9 colourings are fixed by the identity $i$. 
Three of the elements of $S_3$ are transpositions. Only 1 colouring is fixed by any given transposition.
Two of the elements of $S_3$ are cycles of order $3$, and they don't fix any colouring. 
The average number of colourings fixed by the elements of $S_3$ is $(9+1+1+1+0+0)/6$, which is $2$. 
Likewise, you can count the colourings whose column totals are (3,3,0,0), (3,1,1,1), (2,2,2,0) and (2,2,1,1) to get the total number of colourings
[the answer, BTW, is 27]
