9. How many ways can you place a White Knight and a Black Knight on an $8 \times 8$ chessboard such that they do not attack each other?
(A) 1680;
(B) 1712 ;
(C) 3696 ; (1
(D) 3760 ;
(E) None of the above,
9. Answer: (C).
We consider the position of the Black Knight. The number of positions being attacked by the White Knight can be counted.
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline 2 & 3 & 4 & 4 & 4 & 4 & 3 & 2 \\
\hline 3 & 4 & 6 & 6 & 6 & 6 & 4 & 3 \\
\hline 4 & 6 & 8 & 8 & 8 & 8 & 6 & 4 \\
\hline 4 & 6 & 8 & 8 & 8 & 8 & 6 & 4 \\
\hline 4 & 6 & 8 & 8 & 8 & 8 & 6 & 4 \\
\hline 4 & 6 & 8 & 8 & 8 & 8 & 6 & 4 \\
\hline 3 & 4 & 6 & 6 & 6 & 6 & 4 & 3 \\
\hline 2 & 3 & 4 & 4 & 4 & 4 & 3 & 2 \\
\hline
\end{tabular}

There are $16 \times 8+16 \times 6+20 \times 4+8 \times 3+4 \times 2=336$ cases. Hence, the total number of cases that the Kights do not attack each other is $64 \times 63-336=3696$.