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For a finite triangulation of a surface of the sphere, Euler formula for a sphere iswhere $V$, $E$ and $F$ count vertices, edges and triangular faces. Since each face has three edges and each edge borders two faces, also $3F=2E$.

Let $p$ and $h$ count the pentagonal and hexagonal faces. Counting incidences of edges with vertices and faces givesThe Euler formula for a sphere also applies to this polygonal decomposition: triangulating a polygon by adding one diagonal increases both $E$ and $F$ by one, preserving $V-E+F$. Substitution givesThus there are exactly 12 pentagons, independently of the number of hexagons.