I have the follwing combinatorial question, which I motivate below. Let $N:=\lbrace 1,\dots,n \rbrace$ be a set and let $\lambda_1,\dots,\lambda_l$ be a collection of $l:=n+3$ subsets of $N$ with the following properties:

$(1)\ \bigcup \lambda_i=N$

$(2)\ \lambda_i \neq \lambda_j $ for $i\neq j$

$(3)\ |\lambda_i|=4$

$(4)$ Each $\lambda_i$ consists of $2$ unordered pairs of $2$ elements each (Since this is not super precise, I give an example: instead of writing $\lambda_i=\lbrace 1,2,3,4 \rbrace$, we write $\lambda_i=\lbrace 1,2 | 3,4 \rbrace$ to point out the structure of pairs. Moreover, everything is unordered, i.e. $\lbrace 1,2 | 3,4 \rbrace=\lbrace 3,4 | 1,2 \rbrace=\lbrace 2,1 | 3,4 \rbrace=\dots$ and combinations of these)

$(5)$ The set of all $\lambda_i$ is totally ordered, say $\lambda_1>\lambda_2>\dots$

The question I want to ask is motivated by the study of graphs and the resolution of their vertices, so lets add graphs:

Start with a graph consisting of a single vertex $v$ and attach $n$ ends (same $n$ as above, ends=unbounded edges) to this vertex. Label the edges with different numbers from $1$ to $n$.

A *resolution* of $v$ is the following: replace $v$ by two vertices $v_1,v_2$ that are connected by an edge of some finite length and distribute the labeled ends of $v$ among $v_1,v_2$.

Define $\lambda_{v_1}:=\lbrace \lambda_i=\lbrace \beta_1,\beta_2|\beta_3,\beta_4 \rbrace|$ the shortest path between the ends labeled $\beta_1, \beta_2$ intersects the shortest path between the ends labeled $\beta_3, \beta_4$ in $v_1 \rbrace$ and define $\lambda_{v_2} $ in the same way.

Now let's bring the sets $\lambda_i$ into play: A *resolution of $v$ according to $\lambda_1=\lbrace \beta_1,\beta_2|\beta_3,\beta_4 \rbrace$* is a resolution of $v$ such that ends labeled $\beta_1,\beta_2$ are attached to $v_1$ (or $v_2$) and ends labeled $\beta_3,\beta_4$ are attached to $v_2$ (or $v_1$) AND such that (for $t=1,2$) $v_t$ is adjacent to $3+\lambda_{v_t}$ edges.

QUESTION: Resolve $v$ according to $\lambda_1$, then one of the two new vertices (say $v_1$) has the property that $\lambda_2\in\lambda_{v_1}$. Resolve $v_1$ according to $\lambda_2$ and continue like this (with respect to the total order, see $(5)$) until there are no more sets. Doing it right, we should end up with a tree, where every vertex is adjacent to exactly $3$ edges. HOW MANY of these trees are there? Or in other words: How many ways of *totally* resolving a vertex are there?

Here is some motivation: I studied degenerations of tropical curves and want to know how many preimages a degeneration map has. At the moment I have no intuition for these numbers, my first guess was that one should consider how the $\lambda_i$ "overlap", but I have no results or any kind of promising strategy, yet. I am also not aware of any related results, but maybe you are?