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 ($3!$).
$$8\cdot 7\cdot 6 -[6\cdot 4\cdot 2\cdot 3!-2\cdot 4\cdot 3!]=96$$
 A: 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 $8(4\times2+1\times4)=96$.
A: The original version of the question specified that the pieces to place were a horse, a bishop, and a rook. According to a comment from OP, specifying the types was merely to make the pieces distinguishable. I edited the question to make remove the misleading terminology, and I'll leave this answer for posterity. 

Why specify the types of pieces if we're not supposed to take their attack modes into account? 


*

*Horses (aka, Knights) attack in an "L"-shaped manner (eg, two squares up, one square over), which moves them from a white square to a black square. Since no pieces are on black squares, this attack is moot.

*Rooks attack horizontally or vertically. Since no pieces are on the same row or column, this attack is also moot.

*Bishops attack diagonally. This attack remains viable, and dramatically reduces the possible placement of the other pieces.


There are 8 locations of the bishop, each of which limits to possible locations of the horse and rook. Here they are, grouped by symmetry:








As we see, only the first two bishop placements allow the horse and rook to be placed at all, and those placements are interchangeable. (We can't place pieces in the last four, since they'd be on the same row or column.)

Thus, the total number of attack-free ways to place the pieces is: $4$.

A: Paint the white spaces blue (in the first and third columns) and red (in the second and fourth): 

The rule against having tokens in the same row/column implies that each "color zone" has at most two tokens. On the other hand, with three tokens and only two zones, the Pigeonhole Principle requires that at least two tokens go into one zone. Thus, exactly two tokens go into one zone, leaving one token to go into the other zone.
Note that the two tokens in the same color zone must be placed in opposite corners of that zone. The token in the other color zone is unaffected by the placement of the first two tokens.
Now, first considering the tokens indistinguishable, we can count thusly:


*

*Which color zone has two tokens? $2$ choices.

*How to place those two tokens? $2$ ways.

*Where to place the third token? $4$ places.


and then 


*

*How to make the tokens distinguishable? $3!$ ways.



Therefore the total number of placements of distinguishable tokens is
  $$2 \cdot 2 \cdot 4 \cdot 3! = 96$$

