Let's do this piece by piece.
First, let's consider the first rook, we can place it anywhere on the board, thus we have $8^2=64$ choices for that.
Now, for the second one, we can't be in the row or column of that first one, so leaving us with $7^2=49$ choices.
Then so on, we have $6^2=36$ for the third one, $25$ for the fourth one, and so on $\dots$
But, however, we have to remember the rooks are not labeled, thus it doesn't matter specifically about a specific rook's position.
Thus, we have a total of $\frac{(8!)^2}{8!}=40320$ ways.
First imagine that the rooks have student numbers, or that one is black and the other is white.
There are $64$ ways to place the black rook. For each such way, there are $14$ ways to place the white rook, for a total of $(64)(14)$.
But the rooks are probably intended to be identical. Thus the number of black rook/white rook placements must be divided by $2$, for a total of $\frac{(64)(14)}{2}$.
Best Answer
We assume the chessboard is oriented. So there are two players $A$ and $B$ sitting at the board. Then the calculation is simple. We must put a rook in each row. There are $n$ locations for a rook on the top row.
For each such location, there are $n-1$ possible locations for the rook in the second row. Once we have chosen the locations for the rools in the top two rows, there are $n-2$ allowed locations for the rook on the third row, and so on, for a total of $n!$ ways.
Things get much more complicated if we consider two arrangements that look the same if the two players trade places to be the same.