How many ways are they to place 3 non-attacking bishops on an $n \times n$ board such that $2$ of these bishops are placed within the $(n-1) \times (n-1)$ board and the other 1 is placed outside of this sub-board?
See the picture below for an example of the $6 \times 6$ board. The question is how many ways can I place 2 non-attacking bishops into the yellow-colored $5 \times 5$ sub-board and 1 non-attacking bishop into the remaining (white) cells? (all 3 bishops are mutually non-attacking)
I realised some patterns in the sub-sub-board:
A bishop on the cell labelled $3$ in the sub-board will occupy $3$ other white cells. Similarly, a bishop placed in a cell labelled $1$ (as seen in Example) will occupy only 1 other white cell. I also noticed that the number of $3$-labelled cells is the $(n-2)^{th}$ triangular number and for $1$s, it is the $(n-1)^{th}$ triangular number.
Other than this, I am stuck and am not sure how to continue... Can anyone enlighten me? (Note I am interested in the general solution in terms of $n$.)