Question:
If $A$ is a doubly stochastic matrix(https://en.wikipedia.org/wiki/Doubly_stochastic_matrix) of order $n$ whose diagonals entries are all $0$ and each off-diagonal entry is $1/(n-1)$ then can we find some pattern in permanent of $A$ as it is some special kind of matrix?
Thanks in advance!
You have that :
$$ \mbox{perm}(A) = \sum_{\sigma \in S_n} \prod_{i = 1}^n A_{i, \sigma(i)} $$
The only $\sigma \in S_n$ such that the product is not $0$ are those such that $\sigma(i) \neq i$ for all $i$. Denote this subset of permutations $D_n \subset S_n$. For $\sigma \in D_n$, the product is always $(n-1)^{-n}$ so it does not depend on $\sigma$. Therefore :
$$ \mbox{perm}(A) = \frac{|D_n|}{(n-1)^n} $$
I don't think there is an explicit formula for $|D_n|$, except the following (not entirely satisfying) :
$$ |D_n| = n! \sum_{k = 0}^n \frac{(-1)^k}{k!} $$
from which you have the asymptotic behaviour :
$$ \mbox{perm}(A) \sim \sqrt{2 \pi n} e^{-n} $$
as $n \to \infty$.