= Solution
Let $p$ and $h$ count the pentagonal and hexagonal faces. Counting incidences of edges with vertices and faces gives
$$
3V=2E,\qquad 5p+6h=2E,\qquad F=p+h.
$$
The <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 gives
$$
2=\frac{2E}{3}-E+p+h=p+h-\frac{5p+6h}{6}=\frac p6.
$$
Thus \b[there are exactly 12 pentagons], independently of the number of hexagons.
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