Stop nagging! Of course we can apply Euler Characteristics to find out the number of hexagons and pentagons on your football.
For every surface S there exists an integer chi(S) such that whenever a graph G with V vertices and E edges is embedded in it so that there are F faces (regions divided by the graph), we have V - E + F= chi(S).
For a sphere chi(S)=2.
So, if we have P pentagons, and H hexagons we can calculate the number of vertices as follows:
6 vertices for each of the hexagons, i.e. 6*H.
5 vertices for each of the pentagons, i.e. 5*P.
But we have counted each vertex three times (once for each adjacent polygon) so the number of edges E = (6*H + 5*P)/2.
Substituting and solving this equation we get P = 12. So, there are 12 pentagons.
Moving swiftly along, we can see that each pentagon is surrounded by 5 Hexagons. So there should 5*P hexagons, but we have triple counted each hexagon for each of its 3 adjacent pentagons. Thus the number of hexagons is 5*P/3 which equals 20.
That is why - dunderhead - there are 20 hexagons and 12 pentagons in that ball you are standing there bouncing.
Anyway, you are getting under my feet now. Go out and kick 32 polygons worth of leather and air around. I will call you back in when your tea is ready.