Q. 2.17 - A First Course in Probability

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.

Step-by-Step Solution

Answer

The probability is $9.10946\times {10}^{-6}$

Step 1: Count total ways to place 8 rooks

There are 64 squares on a chessboard. The total number of ways to place 8 rooks on the board (unordered) is \(\binom{64}{8}\).

Step 2: Count favorable arrangements

For no rook to capture another, each row and each column must contain exactly one rook.
Choose which column the rook in row 1 goes: 8 choices.
Choose which column the rook in row 2 goes (not the same column): 7 choices.
Continue: the number of favorable arrangements is \(8! = 40320\).

Step 3: Calculate the probability

\(P = \frac{8!}{\binom{64}{8}} = \frac{40320}{4426165368}\)
Let us compute: \(\binom{64}{8} = \frac{64!}{8! \cdot 56!} = 4426165368\).
\(P = \frac{40320}{4426165368} \approx 9.109 \times 10^{-6}\)

Step 4: Final answer

The probability that none of the 8 rooks can capture any of the others is \(\frac{8!}{\binom{64}{8}} = \frac{40320}{4426165368} \approx 9.11 \times 10^{-6}\).