Diagonal rules
How many ways can you fill a standard 8x8 chess board with black and white pawns such that no black pawn is diagonally adjacent to another black pawn? And no white pawn is diagonally adjacent to a white pawn?
Assumption : You have an infinite number of both black and white pawns. And pawns of a particular color are indistinguishable. All squares must contain a pawn. (64 pawns in all)
Clarification : If a board arrangement can be rotated so that it is identical to another, this would count as two different arrangements.
The answer is 4.
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1 solution
The top left corner can be filled with either a black or a white pawn and the square just to the right of it can also be filled with a white or a black pawn.
Once this choice has been made the colors for the pawns on the remaining squares is determined.
So there are 2 × 2 = 4 ways