The number of arrangements of eight rooks on a chessboard such that no two attack each other is $8!$. A way to see this is to note that a placement of a rook is the same as a pair $(i,j)\in \times$. We want to place rooks at different places, and there are $(8!)^2$ ways to do this, and we must divide by $8!$ because rooks are indistinguishable.
Suppose I want to check the number of such arrangement which are all off the main diagonal. I read on this site that these are related to things called derangements, and that there's a magic formula for them involving $e$. I would like to understand what I'm missing in the following naive approach:
First, let's look at the problem in a different way. Instead of thinking about eight identical rooks on different rows of a chessboard, think of eight different objects - the row number of a rook corresponds to the identifier of the object. Then we just want to arrange eight different objects in a row (different columns), so $8!$ objects.
Now to have off-diagonal arrangements, I think we want fixed-point-free permutations (derangements). So object 1 (they have labels) can go to 7 places, object 2 can go to 6 places and so on. This yields $7!$ as an answer. What am I missing?