TITLE: How many ways are there to fill a 3 × 3 grid with 0s and 1s?
QUESTION [24 upvotes]: Extra conditions that put a formal solution out of my reach: the centre cell must contain a $0$, and two grids are equal if they have a symmetry, e.g.
$$\left( \begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 1 & 1\end{array} \right)
=
\left( \begin{array}{ccc}
0 & 1 & 1 \\
1 & 0 & 1 \\
0 & 0 & 0 \end{array} \right)
=
\left( \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 0 \\
1 & 1 & 0\end{array} \right)
$$
For context, this question is part of an investigation into the number of possible checkmate patterns in chess.

REPLY [47 votes]: Use Burnside's lemma. The number of symmetries of the matrix is eight:

the identity, leaving $2^8$ admissible matrices unchanged (the centre cell being fixed)
two 90° rotations leaving $2^2$ matrices unchanged each
a 180° rotation leaving $2^4$ matrices unchanged
four reflections leaving $2^5$ matrices unchanged each

So the number of possible matrices up to symmetry is
$$\frac{2^8+2\cdot2^2+2^4+4\cdot2^5}8=32+1+2+16=51$$