= Solution
{c}
The <Burnside lemma> states that the number of <orbits> of a finite <group action> is
$$
|G|^{-1}\sum_{g\in G}|\operatorname{Fix}(g)|.
$$
For $m=2n-1\geq3$, the <dihedral group> consists of $m$ rotations $r^j$ by $2\pi j/m$, and $m$ reflections $r^js$, each across the line through one vertex and the midpoint of the opposite edge. Thus $|G|=2m=4n-2$.
For the pentagon, the identity fixes $v^5e^5$ colourings. Each of the four nonidentity rotations has a single cycle on <vertices> and a single cycle on <edges>, fixing $ve$ colourings. Each of the five reflections has one fixed vertex and two vertex pairs, and one fixed edge and two edge pairs, fixing $v^3e^3$ colourings. Applying the <Burnside lemma> to simultaneous vertex and edge colourings gives
$$
\boxed{\frac{v^5e^5+4ve+5v^3e^3}{10}.}
$$
Vertex colours and edge colours may be chosen independently; each symmetry must preserve both assignments.
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