A digraph is a graph where every edge is directed. How many digraphs on $n$ vertices are there? So far I have that between any two vertices (say $j$ and $k$) there are 3 options.


*

*there is no edge between $j$ and $k$

*there is an edge directed from $j$ to $k$

*there is an edge directed from $k$ to $j$


And so from this, we can see that there must be $3^{\binom{n}{2}}$ digraphs on $n$ vertices. 
I'm not sure if it should be $\binom{n}{2}$ or $\binom{n}{3}$? Can anybody clarify this for me?

In a directed graph, there are $n(n-1)$ distinct edges, each of these can be present or not. Thus the number of digraphs with $n$ vertices should be
$$2^{n(n-1)}$$
Note that this allows two vertices to be connected by directed edges in both ways. This coincides with the definition on wikipedia (see the example image).
If instead you want to probhibit situations like $v\leftrightarrow v'$ (double-connectedness), you are correct in your enumeration, we pick two vertices in $\binom n2$ ways and assign one of three cases to it independently of the others:
$$3^{\binom n2} = 3^{\frac12 n(n-1)}$$
If the same method was used do allow the fourth case, we would get the original result:
$$4^{\binom n2} = (2^2)^{\frac12 n(n-1)} = 2^{n(n-1)}$$
