TITLE: How many ways to place three distinguishable tokens on the white spaces of a $4$-by-$4$ chess board?
QUESTION [5 upvotes]: 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 ($3!$).
$$8\cdot 7\cdot 6 -[6\cdot 4\cdot 2\cdot 3!-2\cdot 4\cdot 3!]=96$$

REPLY [3 votes]: 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
  $$2 \cdot 2 \cdot 4 \cdot 3! = 96$$