Consider the following directed graph: The number of different topological orderings of the vertices of the graph is _________.

In a topological ordering of a Directed Acyclic Graph (DAG), for every directed edge $u \to v$, vertex $u$ must appear before vertex $v$.

1. Vertex $a$ has in-degree $0$, so it must always appear first (at position 1).

2. Vertex $f$ has out-degree $0$, so it must always appear last (at position 6).

3. The remaining vertices are along two independent paths:

   • Path 1: $b \to c$ (meaning $b$ must appear before $c$)

   • Path 2: $d \to e$ (meaning $d$ must appear before $e$)

4. The 4 intermediate positions (positions 2, 3, 4, 5) must be filled by the set of vertices $\{b, c, d, e\}$.

The total number of valid topological orderings corresponds to the number of ways to interleave the sequences $(b, c)$ and $(d, e)$ into 4 positions:

The 6 possible topological orderings are:

• $a, b, c, d, e, f$

• $a, b, d, c, e, f$

• $a, b, d, e, c, f$

• $a, d, b, c, e, f$

• $a, d, b, e, c, f$

• $a, d, e, b, c, f$

Hence, the correct answer is 6.