# The relation between face counts and edge counts in a polyhedron, $3f_3 + 4f_4 + 5f_5 +\dots = 2E$

Why does $3f_3 + 4f_4 + 5f_5 + \dots = 2E$ hold for every polyhedron?

Notation: $f_k$ is the number of faces with $k$ edges; $E$ is the total number of edges.

Is there a specific proof for this or is it just due to the fact that each edge makes up parts of 2 different faces. I have looked all over the internet and I cannot find a proof online, everyone just seems to 'assume' that it does.

It's a counting argument. Imagine holding your polyhedron. For each $f_3$, put a mark on its 3 edges, for each $f_4$, put a mark on its 4 edges, etc. When you've finished, you've make $3f_3 + 4f_4 + 5f_5 + \cdots$ marks in total (LHS). But each edge has two marks on it (one made from each adjoining face), so another way of counting the marks is $2E$ (RHS).