The probability that two queens, placed at random on a chess board, do not take on each other is $\frac{p}{q}$ (where $H.C.F(p, q) = 1$) then $q - p - 5 = \ldots\ldots\ldots\ldots$
Step-by-Step Solution
Key Concept: Calculate attacking queen positions using rows, columns, and diagonals, then subtract from total arrangements.
Two queens attack each other if placed on the same row, column, or diagonal. The total ways to place 2 queens on a chessboard is $\binom{64}{2}$. The favorable placements (not attacking) equals total minus attacking positions. Attacking positions include 8 rows with $\binom{8}{2}$ each, 8 columns with $\binom{8}{2}$ each, and diagonal arrangements totaling $\binom{7}{2} + 4\left(\binom{6}{2} + \binom{5}{2} + \binom{4}{2} + \binom{3}{2} + \binom{2}{2}\right)$. This gives $P(A) = 1 - \frac{13}{36} = \frac{23}{36}$.
Correct Answer: 8