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 $f(G)=m$?



Are all numbers from $1$ to $n!$ the number of perfect matchings of some bipartite graph?
**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 


