Suppose that two chessboards are also considered equivalent (aside from rotational symmetry) if one can be obtained from the other by complementing red and black colors. How many different $2 × 4$ chessboards are there?
The given answer is $\frac{1}{4}(2^8+2^4+0+2^4)$ , but i did not understand the zero in the parentheses. I thought that the expression must have been $\frac{1}{4}(2^8+2^4+2^4+2^4)$ such that $2^8$ came from identity permutation , $2^4$'s came from $180$ degree rotation , vertical and horizontal flippings.
Can you explain the answer given by the book , and say why my answer is wrong ?
The four relevant symmetries are identity, rotation (by $180^\circ$), complementing colours and complement plus rotation. You already know that the number of boards invariant under the first two symmetries are $2^8$ and $2^4$ respectively.
There are no boards invariant under complement alone, hence the $0$. There are $2^4$ boards invariant under complement plus rotation: the top row of squares may be freely chosen, and this determines the bottom row as the complemented rotation of the top. This then leads to the final answer through Burnside's lemma.