Let $\mathcal{T}$ be a decomposition of the two-dimensional sphere into polygonal domains, with every polygon having at least three edges. Let $V, E$, and $F$ denote the numbers of vertices, edges and faces of $\mathcal{T}$, respectively. State Euler's formula. Prove that $2 E \geqslant 3 F$.

Suppose that at least three edges meet at every vertex of $\mathcal{T}$. Let $F_{n}$ be the number of faces of $\mathcal{T}$ that have exactly $n$ edges $(n \geqslant 3)$ and let $V_{m}$ be the number of vertices at which exactly $m$ edges meet $(m \geqslant 3)$. Is it possible for $\mathcal{T}$ to have $V_{3}=F_{3}=0$ ? Justify your answer.

By expressing $6 F-\sum_{n} n F_{n}$ in terms of the $V_{j}$, or otherwise, show that $\mathcal{T}$ has at least four faces that are triangles, quadrilaterals and/or pentagons.