Probability of placing 8 rooks that cannot attack in a chessboard (error in textbook?) My solution would be $\frac{8!}{64 \choose 8}$ as there are 8 places to put the first rook on the first row, then there are 7 places left on the second row etc etc, then divide this by the total number of ways of choosing 64 squares to occupy with 8 rooks.
Looking elsewhere on this StackExchange, I believe this is correct (this post is slightly different as it's re. probabilities, not just combinations)?
However, the textbook has the following answer:

I am bamboozled as I believe the denominator is $64 \choose 7$ and the numerator seems to ignore the fact that it doesn't matter the order which the rooks are placed on each row.

The textbook answer given has a small error - the denominator product should run to $57$. Otherwise they are the same number.
$$\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{56!\,8!}} = \frac{8!\,8!}{\frac{64!}{56!}} = \frac{\prod_1^8{i^2}}{64\cdot 63\cdot 62\cdots 57}$$
