[Barbeau, Polynomials, page 8]
I am trying to understand the equation shaded in the extract below:
Unfortunately the wikipedia entry only has complete boards (all squares allowed)
Now for some handwavy thinking on that equation shaded:
1- if S becomes forbidden, then we have chessboard $C_1$ by definition, and the polynomial for that is $R(C_1,t)$ , also by definition.
2- if S is allowed, then let us put a rook on it. This means the corresponding row and column becomes off-limits for the remaining rooks (since they must be non-attacking) , effectively giving the reduced $(m-1)(n-1)$ chessboard $C_2$ on which we have 1 fewer rook to place. The polynomial for chessboard $C_2$ with the reduced number of rooks is $R(C_2,t)$ by definition.
3- The polynomial $tR(C_2,t)$ then corresponds to a $mn$ chessboard for which the entire row and column of S is shaded (?)
At this point, I am not so sure:
about that last statement (3)
even so, why the two chessboards $C_1$ and $C_2$ are in distinct union to chessboard $C$ (resp. why the 2 polynomials $tR(C_2,t)$ and $R(C_1,t)$ add up to $R(C,t)$ )
IMO this is thinking about it backwards. The forwards reasoning is: either we put a rook on it (and then we delete the row and column because that's easier than forbidding all the squares in the row and column) or we don't put a rook on it (and we mark it as forbidden to ensure that we don't undo that decision later).
Otherwise, points 1 and 2 look fine to me.
I don't even know what you mean by statement (3). What does "shaded" mean?
I would replace statement (3) with
3b - the polynomial $tR(C_2, t)$ then gives the polynomial for the chessboard $C$ when there is a rook on S.
The key observation is that you multiply the expression from point (2) by $t$ as a way of counting the rook.
Because either there is a rook on S or there isn't. The two options are exclusive and exhaustive.