Question

If a rook (castles) are randomly placed on 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 is$9.10946\times {10}^{-6}$

Step-by-step solution

Step 1

There are $64$squares on a chessboard and to place 8 rooks randomly there are $\left(\begin{array}{c}64\\ 8\end{array}\right)$ways.

As none of the rooks should capture each other.

Every row or column should have 1 rook.

for first rook $8$ways, for second $7$ways and so on

Thus, there are localid="1647096061792" $8!$ways.

Therefore, the probability localid="1647096401848" $=\raisebox{1ex}{$8!$}\!\left/ \!\raisebox{-1ex}{$\left(\begin{array}{c}64\\ 8\end{array}\right)$}\right.=9.10946\times {10}^{-6}$