Let $M$ be an $n\times n$ with elements $\in\{0,1\}$.
Using the fact that interchanging any rows or colums of the matrix does not change its permanent, it is possible to show that $Perm(M) > 0$ iff you can interchange rows and columns of $M$ in a way that its main diagonal is filled with $1$'s.
However, in which case can we say that $Perm(M)=1$? It is possible to show that (assuming $M$ has only $1$'s on its main diagonal) if $\exists m\exists k:$ $\{m$-th row or column of the matrix contains at least $2$ $1$'s$\}$ $\land\forall i\in[1;n]:M_{k,i}\ge M_{m,i}$, then $Perm(M)>1$. But are there cases when this condition is false and still $Perm(M)>1$?