Intro to Probability A person randomly places 9 rooks on a 9 by 9 grid (the 8 by 8 case corresponds to a chess board). What is the probability that none of the rooks can capture any of the other rooks? 

I have a good start on the problem, but I think that I am possibly missing a step or something. Any hits would be apprciated. I know that each time you place a rook, you lose spaces that you can place the next rook. It goes 9^2 spaces to 8^2 spaces to 7^2 spaces...down to 1 space at the end. Thats for the chances to satisfy the problem. For total options it goes 81 spaces to 80 spaces down to 72 when you place the last rook. So my answer is (81*64*49*36*25*16*9*4*1)/(81*80*79*78*77*76*75*74*73*72) Am I missing anything here?Thank you
If you think about it, placing 9 rooks on a 9x9 chessboard so that they don't attack each other is equivalent to asking how many permutations there are of the numbers $1,2,\cdots,9$. This is because each allowed rook configuration corresponds to a $9\times 9$ permutation matrix $M$, with $M_{ij}=1$ if there's a rook at $(i,j)$ and 0 otherwise: there cannot be two 1's on the same row or column. So the answer is immediately $9!/\binom{81}{9}$.
Don't have enough points for a comment, but you seem to be contradicting yourself when you first say that for the first rook you have 9^2 , for the second you have 8^2 and then for the total options you cite 81, 80, etc.