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: Count attacking positions by rows (8×C(8,2)), columns (8×C(8,2)), and diagonals (2×7 + 2×6 + 2×5 + 2×4 + 2×3 + 2×2 + 2×1), then subtract from total placements C(64,2) = 2016 to find non-attacking probability.
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