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.
 A: I suspect there is a small typo in the textbook. If you write out ${64\choose 8}=\frac{64!}{(8!)(56!)}$ and then cancel the $56!$ in the denominator from the numerator, you get $\frac{64\cdot63\cdots57}{8!}$; inverting this and multiplying by $8!$ (that is, dividing $8!$ by this) then gives $\frac{(8!)^2}{64\cdot63\cdots57}$ as the probability. But the numerator here can be written as $\prod_{i=1}^8i^2$, by commuting the multiplications. In other words, the book's solution, modulo that typo, is the same as yours.
As has been noted in the comments, these two answers can be interpreted distinctly combinatorially; what you've written — $\dfrac{8!}{64\choose 8}$ is the number of arrangements of 8 (unordered) rooks in non-attacking positions on the chessboard — that is, just the number of permutations of 8 objects, one per row — divided by the total number of arrangements of unordered rooks on the board. Similarly, the book's answer of $\dfrac{\prod_{i=1}^8i^2}{64\cdot63\cdots57}$ is the number of arrangement of 8 ordered rooks in non-attacking positions divided by the total number of arrangements of ordered rooks on the board; the numerator here can be thought of as placing rook 1 on any of the $8\times8$ squares on the board, then placing rook 2 on any of the $7\times7$ squares that aren't attacked by rook 1, etc. Another interpretation is that it's the way of arranging unordered rooks in non-attacking configurations (i.e., the $8!$ permutations from your numerator) times the number of ways of ordering those rooks (also $8!$).
A: The textbook answer given has a small error - the denominator product should run to $57$. Otherwise they are the same number.
$$\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{56!\,8!}} = \frac{8!\,8!}{\frac{64!}{56!}} = \frac{\prod_1^8{i^2}}{64\cdot 63\cdot 62\cdots 57}$$
