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

$\frac{8!}{\left(\begin{array}{c}64\\ 8\end{array}\right)}$

Step-by-step solution

Given Information.

Let$8$rooks (castles) be randomly placed on a chessboard,

Explanation.

We go row by row.

First, we choose a position in a row $1$from the $8$columns for keeping the first rooks in $\left(\begin{array}{l}8\\ 1\end{array}\right)$the way.

Next, we choose a position in a row $2$from the available $7$columns(one column has been occupied by a rook$1$). This gives us$\left(\begin{array}{l}7\\ 1\end{array}\right)$ choices.

Continuing like this, we get $8!$ways to put the $8$rooks on the chessboard given the conditions.

Total ways of keeping $8$rooks on a chessboard are simple$\left(\begin{array}{c}64\\ 8\end{array}\right)$.

Explanation.

$\frac{8!}{\left(\begin{array}{c}64\\ 8\end{array}\right)}$