On a 7x7 checker board two boxes are painted yellow and the rest black. Board is magical if one can be obtained from the other by applying rotation 
Two of the squares of a 7x7 checkerboard are painted yellow and the rest are black. two color schemes are magical if one can be obtained from one the other by applying a rotation in the plane of the board. how are many inequivalent color schemes possible?

For starters, there are 49 boxes in total and there are 49C2 ways of picking the two yellow boxes which are 1176 ways
clearly, there are far less than 1176 "magical" set ups
At first i attempted to brute force this problem but realized that it was too hard to do that 
according to a source there are $\frac{1176-24}{4}+\frac{24}{2}$ in which i do not know how this was derived and would like some assistance in this problem
 A: Burnside's lemma can be applied. Let $X$ be the set of all possible boards where two squares are colored yellow (or think only about the two points at the centers of the squares). And let $G$ be the symmetry group of a square, that is the rotations by $90, 180, 270$ and $360$ ($=0$, so identity) degrees around the center of the board (since this is what is means to "rotate the board").
The number of orbits gives the amount of inequivalent boards. By Burnside's lemma
$$|X/G| = \frac{1}{4}\sum_{g\in G} |X^g|$$
where $X^g$ is the set of boards (two points at square centers configurations) fixed by each rotation. Denote these $X^0, X^{90}, X^{180}$ and $X^{270}$.
The identity fixes everything so $X^{0} = |X| = 1176$.
The rotation by $90$ degrees can't fix anything: two points can't be rotations by $90$ of each other (otherwise rotating twice would amount to $0$ degrees, but $2*90=180 \neq 360$). Similarly rotation by $270$ ($=-90$) degrees can't fix any boards.
The boards fixed by the 180 degrees rotation are the ones where points are opposite to each other with respect to the center. There are $\frac{7^2-1}{2}=24$ of these. (This is calculated like this: Remove the center, since the center can't be painted if we have 180 rotation symmetry: the other point doesn't have a symmetry point. But every other point has a symmetry point and because it doesn't matter in which order we paint the squares we must halve to get the amount of boards).
So the answer is $\frac{1}{4}\sum_{g\in G} |X^g| = \frac{1176}{4} +  \frac{0}{4}+ \frac{24}{4}+ \frac{0}{4} = 300$. This is the same numerical answer as your formula, after simplification.
A: I just wanted to post a solution anyway as I took a little time to solve this problem as well.
So this was a problem on the AIME, 1997 AIME problem 7.
Yes, as you said, there are $49\choose2$ = $1176$ ways to choose $2$ yellow coloured pieces on any given board. We see that the answer to this question,$\frac{1176−24}{4}+\frac{24}{2}=300$, is almost $4$ times less than $1176$. This is largely because even though most combinations of green do have $3$ other equivalent combinations, the combination pairs that are reflected through the centre point does not, and they only have $1$ equivalent pair. We call these 'special' combinations.

We see that every combination that contains the centre piece cannot be the special combinations we talk about that only have $1$ equivalent pair. For every other piece, we can find another piece, one unique piece, that is the point reflected through the center.  Using this knowledge, we see that there are $\frac{49-1}{2}$ = $24$ pairs of 'special' combinations. 
Thus, there are $1176-24=1152$ 'non-special' combinations. In these combinations, each one reflects to 3 distinct 'equivalent' combinations. From here, we have $\frac{1152}{4}=288$ different 'inequivalent' colour schemes. From the $24$ special combinations, we have $\frac{24}{2}=12$ 'inequivalent' color schemes. Summing those up yields the desired answer.
A: The formula was derived by studying the size of the magical groups. Most of them contain four boards each (that's where the $\frac{\hphantom x}{4}$ comes from). However, some of them do not (that's where ${}-24$ comes from). They are just pairs (which explains $\frac{24}{2}$).
A: It helps to compute  the cycle index $Z(B)$ of the  action of $C_4$ on
the squares of the board. We may then continue either with Burnside or
PET. The cycle index is by inspection given by
$$Z(B) = \frac{1}{4} (a_1^{49} + a_1 a_2^{24} + 2 a_1 a_4^{12}).$$
These are  in order, the identity,  the rotation by $180$  degrees and
the  rotations by  $90$ and  by $270$  degrees. Now  for Burnside  the
colors must be constant on the  cycles.  All colorings are constant on
the cycles of the identity permutation,  which are fixed points, for a
contribution  of $${49\choose  2}.$$ For  the second  type we  need to
choose the  color of the  fixed point  which cannot be  yellow because
that  leaves  one yellow  square  but  the  remaining cycles  are  all
two-cycles. So it  must be black. This leaves us  to choose the yellow
two-cycle  from   the  $24$  possibilities,  for   a  contribution  of
$${24\choose 1}.$$  For the  last cycle  type if  we choose  the fixed
point to  be yellow we have  one yellow square left  but the remaining
cycles are  all four-cycles  so we cannot  place the  remaining yellow
square. Hence  it must be  black. But now  we have two  yellow squares
that   do    not   suffice   to    color   one   of    the   remaining
four-cycles. Therefore this type of  symmetry does not contribute. The
answer is therefore
$$\frac{1}{4}\left({49\choose 2} + {24\choose 1}\right) = 300,$$
which agrees with the OP and the  earlier answer. Now that we have the
cycle index we  may also apply the Polya Enumeration  Theorem (PET) to
get a  complete classification  of colorings with  at most  two colors
under rotational symmetry. We find
$$Z(B)(Y+B) =
\frac{1}{4} ((Y+B)^{49} + (Y+B) (Y^2+B^2)^{24}
+ 2 (Y+B) (Y^4+B^4)^{12}).$$
Coefficient  extraction  may be  done  manually  which duplicates  the
Burnside argument. A CAS tells us that the coefficients are
$${B}^{49}+13\,{B}^{48}Y+300\,{B}^{47}{Y}^{2}+4612\,{B}^{46}{Y}^{3}
\\ +53044\,{B}^{45}{Y}^{4}+476796\,{B}^{44}{Y}^{5}
+3496460\,{B}^{43}{Y}^{6}+\cdots$$
and we have confirmation of the result from Burnside.
