Let $U = \{1, 2, 3\}$. Let $2^U$ denote the powerset of $U$. Consider an undirected graph $G$ whose vertex set is $2^U$. For any $A, B \in 2^U$, $(A, B)$ is an edge in $G$ if and only if (i) $A \neq B$, and (ii) either $A \subsetneq B$ or $B \subsetneq A$. For any vertex $A$ in $G$, the set of all possible orderings in which the vertices of $G$ can be visited in a Breadth First Search (BFS) starting from $A$ is denoted by $\mathcal{B}(A)$. If $\emptyset$ denotes the empty set, then the cardinality of $\mathcal{B}(\emptyset)$ is ________.

The vertex set of the graph $G$ is the power set of $U = \{1, 2, 3\}$, so the total number of vertices is:

An undirected edge exists between any two distinct subsets $A$ and $B$ if and only if one is a proper subset of the other ($A \subsetneq B$ or $B \subsetneq A$).

For the empty set $\emptyset$, it is a proper subset of every other non-empty subset $S \subseteq U$:

Therefore, $\emptyset$ is directly connected to all remaining $2^3 - 1 = 7$ vertices in the graph $G$.

In a Breadth First Search (BFS) starting from $\emptyset$:

1. The starting vertex $\emptyset$ is visited first (1 choice).

2. Since all other 7 vertices are at distance 1 from $\emptyset$, they will all be visited at level 1 in any possible permutation.

Thus, the number of possible BFS orderings starting from $\emptyset$ is: