Q. What is the total number of ways in which four distinguishable pieces be placed on an $8$X$8$ chess board such that no two pieces are in the same row or column?
This question is remarkably similar to the question in which there are eight pieces, but this case is different, as both cannot be solved by the same method.
If there were eight pieces, the number of ways would have been simply $(8!)^2$. That could be thought as one piece per each row, with the piece in the first row having $8$ possible places, the second having $7$ and so on, giving $$ 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot2 \cdot1 = 8!$$ for each row. Multiplying the $8!$ cases for each column, we get $(8!)^2$
I am not able to think of any helpful approach to the four piece case.
How can the four piece case be solved?