Who got hit?

Let $Q_n$ be the number of ways of placing two non-attacking queens on an $n \times n$ board. We know that $Q_1 = Q_2 = 0$ . Suppose now that $n \ge 3$ .

• There are $Q_{n-1}$ ways of placing two non-attacking queens on an $n \times n$ board so that no queen is in either the last row or column,
• There are $2n-1$ squares in either the last row or column, and any queen on one of those squares attacks a total of $3(n-1)$ other squares,
• There are $(n-1)(n-2)$ ways of placing two non-attacking queens, so that each is in either the last row or column.

Thus $Q_n \; = \; Q_{n-1} + (2n-1)\big(n^2 - 3(n-1) - 1\big) - (n-1)(n-2) \; = \; Q_{n-1} + 2n^3 - 8n^2 + 10n - 4 \hspace{2cm} n \ge 3$ and hence $Q_N \; = \; \sum_{n=3}^N(2n^3 - 8n^2 + 10n - 4) \; = \; \boxed{\tfrac16N(N-1)(N-2)(3N-1)}$