Is it possible to build this graph?

In a graph, if we add an edge then the degree of the two incident vertices each increases by one. Thus the sum of the degrees increases by two. As a result, the sum of degrees of of vertices in a graph is always an even number. Now calculate that your list of numbers sums up to an odd number.

Formulas:

Undirected Graph: $\sum_{v \in V} deg (v)$ = $2$$\mid E \mid$

Directed Graph: $\sum_{v \in V} deg^- (v)$ = $\sum_{v \in V} deg ^+(v)$ = $\mid E \mid$

Explanation:

This problem is an application of the Handshaking Lemma, proven by Euler. The lemma states that every finite, undirected graph (which is one possibility of your graph) has an even number of vertices with an odd degree. The formula is as follows: $\sum_{v \in V} deg (v)$ = $2$$\mid E \mid$. Applying this formula, the sum of all your degrees within the provided set would be 31, which is not divisible by two and hence this graph cannot be built with undirected edges.

I am not sure how to consider this problem with directed edges because with directed edges there is no generalized "degree" term. There are in-degrees and out-degrees for each vertex, however I don't know how many are designated to each node so I couldn't specifically answer your question (unless someone else could elaborate). However, the application is similar to the one for undirected edges. In this situation, you would apply the degree-sum formula for digraphs $-$ which is essentially the Handshaking lemma for digraphs. This lemma states:$\sum_{v \in V} deg^- (v)$ = $\sum_{v \in V} deg ^+(v)$ = $\mid E \mid$ meaning that in order for this graph to exist, the sum of the in-degrees must be equal to the sum of the out-degrees which must be equal to the number of edges.