Given a Ferrers board of shape $(b_1,\ldots,b_m)$, we define $r_k$ as number of ways to place $k$ non-attacking rooks (as in Chess). In section 2.4 of Stanley's Enumerative Combinatorics (vol. 1) it's shown the identity: $$\sum_k r_k (x)_{m-k} = \prod_i (x+s_i)$$ where $s_i = b_i-i+1$, but I don't know if I can invert this formula or make an efficient algorithm to compute the $r_k$'s.

If this isn't possible, I would be satisfied if I can compute them efficiently in the following shapes:

$(2,2,4,4,\ldots,2n-2,2n-2,2n)$

$(2,2,4,4,\ldots,2n,2n)$

$(1,1,3,3,\ldots,2n-1,2n-1,2n+1)$