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 total queen placements as C(64,2) = 2016, then subtract attacking positions (same row, column, or diagonal). For each row/column: C(8,2) = 28 placements. For diagonals: count pairs on each of 15 diagonals (lengths 1,2,3,...,8,...,3,2,1) giving 7+12+15+16+15+12+7 = 84 attacking diagonal pairs. Non-attacking probability = (2016-8×28-8×28-84)/2016 = 1232/2016 = 77/126, so p=77, q=126, and q-p-5=8.
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