I know that this topic as been mentioned before, but no accurate answer has been provided.


Suppose we have to place $n$ rooks on $n \times n$ chessboard so that no one attacks another. How to count the number of different ways to place them up to rotations of the chessboard?


I have been trying the Burnside's lemma but did not obtain satisfactory results...



Number of different positions of rooks on chessboard
It seems that the first complete solution to this kind of problem was worked out by Édouard Lucas in section 128 of his textbook on number theory:


* Édouard Lucas, [Théorie des Nombres](https://archive.org/stream/thoriedesnombre00lucagoog#page/n245/mode/2up) (1891)


Notice that he considers solutions different up to rotations *and reflections*, so you have to slightly modify the argument. Here is an outline:


You begin with the $n!$ total solutions, and consider the group of symmetries $\Gamma$ of the chessboard you are interested in. Those are:


$\Gamma\_1$: Identity


$\Gamma\_2$:Rotation by $\pi$ radians.


$\Gamma\_3$:Rotation by $\pm \pi/2$ radians.


If we denote $\sigma\_n$ the number of differente solutions (under those symmetries), then $\sigma\_n$ is the number of orbits of $\Gamma$ on the set of the $n!$ solutions, so that you can now apply Burnside's lemma.


Now, $\Gamma\_1=n!$ and $\Gamma\_2=(n/2)!2^{n/2}$.


Also $\Gamma\_3=(2m)!/m!$ if $n=4m$ or $n=4m+1$, and $\Gamma\_3=0$ otherwise.


From here you can easily count the number of solutions.


For a clearer and more modern exposition of the solution considering also reflections, you can consult:


* Robert W. Robinson, [Counting arrangements of bishops](http://link.springer.com/chapter/10.1007%2FBFb0097382) (1976)


