1. In how many ways can two knights - a white one and a black one - be placed on a chessboard consisting of $16 \times 16$ cells so that they threaten each other? (A knight moves in an "L" shape, i.e., it can move to one of

![](https://cdn.mathpix.com/cropped/2024_05_06_c995faa73b19f5c426fdg-01.jpg?height=217&width=234&top_left_y=274&top_left_x=1659)
the fields closest to the one it is on, but not on the same row, column, or diagonal.) (12 points)
Solution: Let's connect an $n \times n (n>2)$ rectangular coordinate system to the board. Denote the coordinates of the two knights as $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$, where $x_{k}, y_{k} \in\{1,2, \ldots, n\}, k=1,2$. The knights threaten each other if 1) $\left|x_{1}-x_{2}\right|=1, \quad\left|y_{1}-y_{2}\right|=2$ or 2) $\left|x_{1}-x_{2}\right|=2, \quad\left|y_{1}-y_{2}\right|=1$. In the first case, there are four such possibilities:

$1 \leq x_{1} \leq n-1, x_{2}=x_{1}+1, \quad 1 \leq y_{1} \leq n-2, y_{2}=y_{1}+2 ;$

$1 \leq x_{1} \leq n-1, x_{2}=x_{1}+1, \quad 1 \leq y_{2} \leq n-2, y_{1}=y_{2}+2$

$1 \leq x_{2} \leq n-1, x_{1}=x_{2}+1, \quad 1 \leq y_{1} \leq n-2, y_{2}=y_{1}+2$

$1 \leq x_{2} \leq n-1, x_{1}=x_{2}+1, \quad 1 \leq y_{2} \leq n-2, y_{1}=y_{2}+2$.

Each of these sub-cases corresponds to $(n-1)(n-2)$ positions. The second case differs from the first by a $90^{\circ}$ rotation. For $n=16$, we have $8(n-1)(n-2)=8 \cdot 15 \cdot 14=1680$.

Answer: 1680.