Relation between number of edges and the sum of number of vertices of degree k

Question

Let P be a polyhedron and let G be its associated graph. Suppose P has V
vertices, E edges, and F faces. For each k, let $V_k$ be the number of vertices of
degree k, and let $F_k$ be the number of faces of P (or regions of G) of bound degree k. Since every edge of P
touches exactly two vertices and exactly two faces, we find that
$$sum kV_k=2E= sum kF_k$$
I don't understand why that relation is true.

Kenny Wong · Answer

$V_k$ is the number of vertices that are connected to exactly $k$ edges. So $sum kV_k$ is the sum, over all vertices, of the number of edges connected to each vertex. This is the same as the sum, over all edges, of the number of vertices connected to each edge. Since each edge is connected to exactly two vertices, this is equal to $2E$.
$F_k$ is the number of faces that have exactly $k$ edges. So $sum kF_k$ is the sum, over all faces, of the number of edges each face has. This is the same as the sum, over all edges, of the number of faces that contain each edge. Since each edge is contained in exactly two faces, this is equal to $2E$.

Relation between number of edges and the sum of number of vertices of degree k

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