How many ways to place three distinguishable tokens on the white spaces of a $4$-by-$4$ chess board? 

Consider a $4\times 4$ chessboard like the one showing above. You want to place 3 distinguishable tokens. Each piece can only be placed in a white square, with only one piece per row and per column.
How many ways are there to arrange the pieces?

I have one way of solving this, but I really want to know way to solve this "directly".
Solving Indirectly: All the cases ($8\cdot 7\cdot 6$) except the ones that break the rules.
So the cases that break the rules are:
Imagine that I fix two points horizontally that break the rule, then I have 6 places to place the other one ($6$), then I can do this in every row ($4$), I can do this vertically as well ($2$), and then I have to permute the 3 pieces ($3!$), however I will repeat the cases where I can't tell if I'm doing this vertically or horizontally(the intersections of both) so, those cases are, I fix two points and I have 2 places for other one ($2$), and I can do this in every row ($4$), and I don't multiply by two because I only want to eliminate one time, and I permute the 3 pieces ($3!$).
$$8\cdot 7\cdot 6 -[6\cdot 4\cdot 2\cdot 3!-2\cdot 4\cdot 3!]=96$$

You have $8$ possible places to put the pieces. After the first placement, two other places are ruled out due to being in the same row or column, so now you only have $5$ choices. Among these, $4$ choices will eliminate two more choices, and $1$ will eliminate none. Your answer is thus $8(4\times2+1\times4)=96$.
