TITLE: Special formula for the permanent of the sum of two matrices
QUESTION [5 upvotes]: Dear math stack exchange community,
I was told that in the paper
http://www.tandfonline.com/doi/abs/10.1080/03081088708817770
there was a formula for the permanent of the sum of two matrices $X$ and $Y$ via permanents of certain submatrices of $X$ and $Y$.
Unfortunately, I don't have access to this paper (and thus, to this formula) at the moment.
I would be very grateful. if somebody could give the formula here.

REPLY [6 votes]: For all $k\leq n$, let $Q_{k,n}$ the set of all finite sequences of positive integer $\alpha=\{\alpha_{i}\}_{i=1}^{i=k}$ such that $$
1\leq \alpha_1 < \alpha_2< \ldots <\alpha_{k-1}< \alpha_k\leq n.
$$
If $A$ happens to be a $n\times n$ matrix and if $\alpha,\beta\in Q_{k,n}$ then $A(\alpha|\beta)$ denote a $(n-k)\times(n-k)$ submatrix submatrix of $A$ obtained by deleting the rows $\alpha$ and the  columns $\beta$; Let $A[\alpha|\beta]$ the complement of $A(\alpha|\beta)$ in $A$.

THEOREM 1 (Expansion of the   permanent of the sum of two matrices)
Let $A$ and $B$ be arbitrary $m\times m$ matrices.  Then
$$
\mathrm{per}(A+B)=\mathrm{per}(B)+\sum_{k=1}^{m-1}\sum_{\alpha,\beta\in Q_{k,m}}\mathrm{per}A[\alpha|\beta]\cdot\mathrm{per}A(\alpha|\beta)+\mathrm{per}(A)
$$

See also in wikipedia verbete on the permanents and its references.