Number of configurations of coins on a chessboard For each square of a standard 8 × 8 chessboard, you either put a coin on it or leave it empty. You also have to guarantee that each row and column will contain an odd number of coins. How many such configurations of coins are there?
Notice: the chessboard is unlabeled, but the left-bottom-corner is black.
 A: We can think of the chess board as an $8\times 8$ matrix in $F_2$. The number of configurations with an odd number of coins on each row/column is the same as the number of configurations with an even number of coins on each (the bijection is given by adding the identity matrix), and it will be easier to work with the latter.
We want to count the number of matrices that obey the row/column parity condition, up to rotations by 180 degrees. The number is therefore given by
$$(T-S)/2 + S,$$
where $T$ is the total number of matrices satisfying the parity condition, and $S$ is the number of such matrices with 180-degree symmetry.
Claim: Each matrix $M$ with even row/column sums is uniquely specified by $49$ binary digits $\alpha_{i,j}$, with
$$M = \sum_{i=1}^7 \sum_{j=1}^7 \alpha_{i,j} A_{i,j},$$
where $A_{i,j}$ is the matrix that is zero except for a $2\times 2$ block of ones whose top-left corner is at $(i,j)$ (and lower-right corner at $(i+1,j+1)$), and the sum is taken in the space of matrices with coefficients in $F_2$, ie the coefficients are taken mod 2.
Proof: Start at the top-right corner of $M$ and work your way left-to-right, top-to-bottom.
The above gives us $T = 2^{49}$. We can also compute the number of symmetric matrices with even row/column sum: we can pair off each $A_{i,j}$ with the corresponding basis matrix $A_{8-i,8-j}$ that you get by rotating $A_{i,j}$ by 180 degrees. Only one basis matrix gets mapped to itself: $A_{4,4}$. Therefore you have two choices for $\alpha_{4,4}$, and two choices for each pair of remaining coefficients, for a total of $S=2^{25}$ choices.
The answer is therefore $2^{48} + 2^{25} - 2^{24} = 2^{48} + 2^{24}.$
