The Problem:-
"In chess, a rook attacks any piece in the same row or column as the rook, provided no other piece is between them. In how many ways can $8$ rooks be placed on a $[8\times8]$ chessboard so that no two attack each other? What about $8$ rooks on a $10\times10$ board?"
I believe I have an answer for the first part of the question. When placing the first rook, there are 8 places on any particular column (or row) to place the rook, leaving just 7 places on a different column (or row) for the next rook, and so on, providing 8! possible ways to place the rooks in such a way that they cannot attack each other ($P(8,8) = 8!/(8-8)! = 8!$).
However, I am not sure I fully understand how this would work for a board where there are more rows and columns than pieces (such as on a 10×10 board). Does it become $P(10,8) = 10!/(10-8)! = 10!/2$ ? If so, why? If not, how should I approach this problem?
This problem was found in "Introduction to Combinatorics and Graph Theory" by David Guichard.
Best Answer
So for the first question you need to place a rook in each column. The rook in the first column can go in $8$ places, the rook in the second column then has $7$ possible rows etc.
In the second part, you need first to choose the $8$ columns out of the $10$ available, and then place a rook in each of the columns. If once again you move from left to right, counting the number of possibilities, you should be fine.
Rather than talking about "any particular" column or row, you should make a habit of defining the scheme more closely (the first, second rather than "any" etc). This will help to avoid a great deal of confusion and can also highlight any double-counting.