Let $G=(V,E)$ be an undirected graph. I call a valuation of $G$ a function $\nu: V \to E$ that maps every node $x \in V$ to an edge incident to $x$ (so that there are $\prod_{x \in V} d(x)$ valuations of $G$, where $d(x)$ is the degree of node $x$). I say that $\nu$ is satisfying if there exist an edge $e\in E$ such that both endpoints of $e$ are mapped to $e$ by $\nu$. I am interested in the following problem:
INPUT: An undirected graph $G$
OUTPUT: The number of satisfying valuations of $G$
My question: What is the complexity of this problem, and does it already have a name?
My guess is that it is #P-hard, even for bipartite graphs. A closely related #P-hard problem is #POSITIVE-2-DNF, or even [#PARTITIONED-POSITIVE-2-DNF][1]. Indeed, you can see an instance of #(PARTITIONED-)POSITIVE-2-DNF as a (bipartite) graph $G$, and you say that a valuation of $G$ either maps a node $x$ to all of its incident edges or to none of them. So my problem is somewhat a variant of #POSITIVE-2-DNF, but where valuations map variables to a single clause in which they occur, instead of mapping them to $0$ of $1$.
==== UPDATE ====
As a3nm showed in his answer, the problem is hard on 3-regular graphs with multi-edges. My answer shows that the problem is also hard on $2$-$3$ regular simple graphs. There is the minor question of knowing if it is hard one $3-regular simple graphs. I don't really care about it, but I still leave it here for completeness.