Are all numbers from $1$ to $n!$ the number of perfect matchings of some bipartite graph? Let $f(G)$ give the number of perfect matchings of a graph $G$.
Consider set $\mathcal N_{2n}=\{0,1,2,\dots,n!-1,n!\}$.
Consider collection of all $2n$ vertex balanced bipartite graph to be $\mathcal G_{2n}$.
For every $m\in\mathcal N_{2n}$ is there a $G\in \mathcal G_{2n}$ such that $f(G)=m$?
 A: Consider $A$ the order $n$ adjacency matrix which has 1 in the $i,j$th entry if there is an edge in the bipartite graph $G$ from top vertex $i$ to bottom vertex $j$, 0 if not. Then the permanent of $A$ counts the number of perfect matchings in $G$.  One now sees that any such bipartite graph missing $k\lt n$ edges is a multiple of $(n-k)!$, which has some significance when $k$ is small.  In particular when $k=2$, the number of matchings is a multiple of $(n-2)!$. So the feasible values near $n!$ are indeed few in number.
Elsewhere (Number of unique determinants for an NxN (0,1)-matrix) I talk about the determinant spectrum problem for order $n$ $0-1$ matrices; a similar problem for permanents can be and probably has been asked, for which I would like to see references.
Gerhard "Ask Me About Multilinear Functions" Paseman, 2016.05.18.
A: No.  For all $n \geq 3$, there is no $G \in \mathcal{G}_{2n}$ with $f(G)=n!-1$.  To see this, note that $f(K_{n,n})=n!$ and $f(K_{n,n}-e)=n!-(n-1)!$, where $K_{n,n}-e$ is $K_{n,n}$ minus an edge.  Thus, there are no graphs $G$ in $\mathcal{G}_{2n}$ with $n!-(n-1)!+1, n!-(n-1)!+2 \dots$, or $n!-1$ perfect matchings 
