Chess Probability 8 rooks You have 8 rooks. What is the probability of placing all 8 rooks on an 8 by 8 chess board with out one being able to hit each other? But there's a catch of course..one of the spaces is unavailable to be used. That being said, there could potentially be 2 rooks in a certain row or column. The unavailable space is 7 columns over and 7 rows down.
Without the catch it would be $\frac{8!}{64\choose 8}$ 
I came up with.. $\frac{8\cdot{7\choose 2}6\cdot6\cdot5\cdot4\cdot3\cdot2}{63 \choose 8}$ 
but I don't think I'm correct. 
Any advice/ideas?
 A: There are ${64\choose8}$ ways to place the rooks when there are no restrictions. Exactly one eighth of these have a rook on the forbidden square. It follows that there are
$${7\over8}\>{64\choose8}=3\,872\,894\,697$$
admissible placements of the rooks.
There are $8!$ good placements of the rooks when there are no restrictions. Exactly one eigth of these have a rook on the forbidden square, so that 
$${7\over8}\>8!=35\,280$$
good placements remain having no two rooks in one row or column.
We can have two rooks in row$_7$, but no two rooks in the same column. The first rook in row$_7$ can be placed in $6$ ways. Then we can choose the row remaining empty in $7$ ways, and finally we can place the remaining $6$ rooks such that a good configuration results in $6!$ ways.  This makes for
$$6\cdot 7\cdot 6!=30\,240$$
good configurations of this type, and the same number results when we start with col$_7$ having two rooks.
When we have two rooks in row$_7$ as well as in col$_7$ then we can place the first rook in row$_7$ and the upper rook in col$_7$ in $6$ ways each; then we can choose the row and the column remaining empty in $5$ ways each, and finally we can place the remaining $4$ rooks such that a good configuration results in $4!$ ways.  This makes for
$$36\cdot 25\cdot 4!=21\,600$$
good configurations of this type, 
All in all there are $117\,360$ admissible good placements, so that the required probability $P$ computes to
$$P={117\,360\over3\,872\,894\,697}\doteq3.03029\cdot10^{-5}\ .$$
When there are no restrictions the corresponding probability $P_0$
is given by
$$P_0={8!\over{64\choose8}}\doteq9.10947\cdot10^{-6}\ ,$$
which is considerably smaller.
A: The easiest way I see to count successes is to take the number of successes for the full chess board, $8!$ and note that $\frac 18$ of them use the forbidden square.  That gives $7\cdot 7!$ successes.  Another way to see this is to place a rook in the column with the forbidden square, which gives $7$ choices.  Then the next rook also has $7$ choices and so on.  The chance of success is $\frac {7\cdot 7!}{63 \choose 8}=\frac {560}{61474519}\approx 9.11\cdot 10^{-6}$  The probability is identical to the case with a complete board.
