There is a simpler solution. First place a \( 2 \times 3 \) or \( 3 \times 2 \) rectangle, then choose a pair of corners in this rectangle to place the knights. There are \( n - 1 \) choices for the column and \( n - 2 \) choices for the row of a \( 2 \times 3 \) rectangle, and the same number but with the axes flipped for a \( 3 \times 2 \) rectangle. In total, there are \( 2 \times 2 \times (n-1) \times (n - 2) = 4(n-1)(n-2) \) ways to place the knights.

Hi James! Did you read the section Counting Placements from Minimal Attack Sections? The simpler solution you present is exactly the simple solution already discussed in that section.