How many ways can $4$ distinguishable pieces be placed on an $8\times8$ chessboard, such that no $2$ pieces are in the same row or column? Question. How many ways can a 4 distinguishable pieces be placed on an $8\times8$ chessboard, such that no 2 pieces are in the same row or column?
Attempt. I went ahead and gave it a drawing. My first thoughts were starting with a single piece placing it in one of the 64 places, then proceeded with trying to put up the next piece, but realised that now $(64-1)-7\cdot2=49$ places were left for the second one to be placed. Then I went with the 3rd piece, and realised there were $(49-1)-6\cdot 2=36$ places left for the 3rd piece, and $(36-1)-5\cdot 2=25$ left for the 4th final one. This could be transformed into a formula but I don't think it serves us here at all. My final answer was adding all of those 4, and I don't even know if it's right, but I'm trying to think of a combinatorics method with the Choose and Permutations and all of that.
 A: A more systematic approach.
This problem isn't too hard if you think about it logically. For the first piece, you have $64$ choices. After placing this first piece, no matter where you placed it, you have removed $15$ available squares:

This means you have $64-15=49$ choices for the next piece. After you place this piece, you will again remove $15$ available squares, minus the two intersections you will have with previously removed squares:

This means that for the third piece, you have $49-13=36$ options. No matter which of these options you choose, you will remove $15$ available spots minus the $4$ intersections you will have with previously removed squares:

So for the final piece you have $36-11=25$ options.

So our answer is
$$64\cdot 49\cdot 36\cdot 25=2~822~400$$
Which, as others have noted, is equal to
$$4!^2\cdot {{}_8 C_4}^2$$
A: Alternatively, we first choose $4$ rows and $4$ columns which can be done in $\displaystyle {8 \choose 4} {8 \choose 4}$ ways.
Now as the pieces are distinct, we decide their order in rows in $4!$ ways and for each such order, we have $4!$ ways of ordering them in columns.
So total number of arrangements $\displaystyle  = 4! \ 4! {8 \choose 4} {8 \choose 4} $
