Number of non-isomorphic ways the following graph can be labelled In how many non-isomorphic ways can the following graph be labelled?

Ignore the numbers on the graph vertices.
I got two different answers and I'm not sure which one of my reasoning is right:
1) (5C2) ways to choose the 2 deg 3 vertices, 3 ways to choose the vertex of deg 2 connected to the two deg 3 vertices, and 2 ways to choose the two remaining vertices = 60
2) 5 ways to choose the vertex connected to the two vertices of deg 3, (4C2) ways to choose the two vertices of deg 3, and two ways to choose the remaining vertices = 60
3) Or (5C2) ways to choose the two vertices of deg 3, (3C2) ways to choose the two vertices of deg 2 at the bottom = 30
 A: The third way is wrong since it doesn't account for the two possible orderings of the two degree-2 vertices.  The correct answer is $60$.
The way I figure it is to start with all possible labelings ($5! = 120$) and then divide by $2$ to account for the symmetry of reflection.
A: 
In how many non-isomorphic ways can the following graph be labelled?

The answer to this question is $1$.  It's one graph, relabelling the vertices will give an isomorphic graph.
The question should be:

How many labelled graphs are isomorphic to this graph?

Then the answer is given by the Orbit-Stabiliser Theorem.  In this case, we have the symmetric group acting on the vertex labels $\{1,2,3,4,5\}$.  Thus, the size of the orbit (i.e., the number of isomorphic labelled graphs) is $$\frac{n!}{|\mathrm{Aut}(G)|}$$ where $\mathrm{Aut}(G)$ is the automorphism group of the graph.
We observe:


*

*Vertex $5$ is the only vertex that belongs to a $3$-cycle and has degree $2$.  So any automorphism must fix the vertex $5$.

*Vertices $3$ and $4$ have degree $3$ whereas vertices $1$ and $2$ have degree $2$.  Hence the sets $\{1,2\}$ and $\{3,4\}$ are preserved by any automorphism.
This leaves us with four possible automorphisms:


*

*$\mathrm{id}$,

*$(12)$,

*$(34)$, and

*$(12)(34)$.


We can eliminate $(12)$ since there is no edge between $2$ and $3$.  We can eliminate $(34)$ for the same reason.  We can check that $(12)(34)$ is indeed an automorphism of the graph.  Hence we can conclude that $|\mathrm{Aut}(G)|=2$ and hence there are $5!/2=60$ labelled graphs which are isomorphic to this graph.
