Question

If 8 rooks (castles) are randomly placed on a chessboard, compute the probability that none of the rooks can capture any of the others. That is, compute the probability that no row or file contains more than one rook.

Short answer

The probability that no row or file contains more than one rook$9.109\times {10}^{-6}$

Step-by-step solution

Step1

Since rooks attack horizontally & vertically, you can't have a rook in the same row or column as another.

So in the first row, you place a rook. There are $8$possible places for first one. In the next row you place a rook,it can't be in the same column as other rook so for this one we have $7$possibilities i.e. $8!$ways

Now total number of possible ways is$\left(\begin{array}{c}64\\ 8\end{array}\right)$

so the probability that no rooks capturing $=\frac{\left(\begin{array}{c}64\\ 8\end{array}\right)}{8!}$$=9.109\times {10}^{-6}$