What is the probability that when you place 8 towers on a chess-board, none of them can beat the other. What is the probability that when you place 8 towers on a chess-board, none of them can beat the other.
Attempt: ${64 \choose 8}^{-1} \approx1$ in $4\ 400\ 000\ 000$
Correct answer: ${64 \choose 8}^{-1} \cdot 8! \approx 1$ in $9\ 000\ 000$.
I disagree with the $8!$. If there's combinations (binomial coefficient) in the denominator, why would there be permutations i.e. the order counts, in the numerator?
 A: There are $\binom{64}{8}$ ways to place the eight rooks on the board. Out of these, there are $8!$ ways for the rooks not to be able to beat each other.
Why? There must be one rook at each row, one at each column. So the placement will define a one-to-one map from the eight columns to the eight rows. There are $8!$ such maps.
(The “towers” are called rooks in English.)
A: The correct answer is the number of good positions of $8$ towers divided by the number of all positions of $8$ towers. The number of all positions is, of course, $64\choose 8$, as you can pick any $8$ positions and put towers on them.
The question you must answer now is why are there $8!$ good positions? 
A: That there is combination but not permutation in the denominator is supported by the fact that we need to choose 8 position out of the total 64 to place the towers, and since the towers are identical there is no need to consider the 8! combinations of every choosed 8 positions.
Now, as to understand why there is permutation in the numerator, let' look from a different point of view. Whenever you place a tower, you cannot place another in its own row or column. Try visualizing the row and column corresponding to the tower getting shaded as you move the tower across the board. After having placed all the 8 towers, if you look from the side of the chess board, they all appear in a line. There are 8! combinations possible in this line, and here we do not avoid the 8! permutation because though we are visualizing all 8 in a line, on every next combination they will not exchange places. Rather they will exchange their row number but remain in the same column. This way we will cover every row position in all 8 column thus covering the entire chess board. This gives us 8! combinations for the towers to be placed such that none of them can beat the other.
