Q. 2.17 - A First Course in Probability

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.

Step-by-Step Solution

Answer

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

Step 1 Given Information.

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

Step 2 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$\left(\begin{array}{c}64\\ 8\end{array}\right)$.

Step 3 Explanation.

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