TITLE: Permanent of a complete graph with negative cliques
QUESTION [6 upvotes]: Let $K_n$ be the complete graph on $n$ vertices. Inside $K_n$ there are $k$ negative(edge weight is equal to -1) complete subgraphs $K_{n_1}, K_{n_2},...,K_{n_k}$, which are vertex disjoint. Let $A(G)$ denotes the adjacency matrix of graph $G$. What will be the permanent of $A(K_n)$?. With suitable relabeling of vertices we can write 
$A(K_n)=\begin{bmatrix}
-A(K_{n_1}) & \\
& -A(K_{n_2})& & J\\
& & \ddots\\
J ^T& & & A(K_m)
\end{bmatrix},$
where, $m=n-\sum_{i=1}^{k}n_i$ and $J$ denote space filled with all 1.

REPLY [4 votes]: Using John Machacek's notation, which differs from the OP's, application of Ryser's formula gives
$$\mathrm{perm}\; A_{m,n}
  = \sum_{s=0}^m \sum_{t=0}^n (-1)^{s+t} \binom ms\binom nt
    (t-s+1)^s (t-s)^{m-s} (t+s-1)^t (t+s)^{n-t}.$$
It might readily simplify to John's formula; I didn't try.  I don't see how to reduce it to a 1-dimensional sum.
The case with $k$ blocks instead of 2 blocks can be written as a similar $k$-fold sum by the same method.