How many sewings are there on a soccer ball? 
A soccer ball is obtained by sewing $20$ hexagonal pieces of leather and $12$ pieces of leather of pentagonal shape.
A sewing joins together the sides of two adjacent pieces. How many sewings are there ?


My effort 
I was able to solve this problem by realizing that if I count the number of sewings adjacent to the hexagons and the ones adjacent to the pentagons I will be counting each sewing twice.
So, the number of sewings is $$\cfrac{120 + 60}{2}=90.$$

Second Approach (this is the one I am asking about)

If we count the sewings adjacent to the pentagons we have $12 \cdot 5 =60 $ sewings ,now to count the rest of the sewings I just observe that any other sewing starts at the edge of some pentagon,so I have $60$ other sewings,for a total of $120$ sewings .
However this doesn't quite work, but if I look at the picture I have posted above it seems to be correct as I don't have any pentagon sharing a sewing with another pentagon.
What am I missing?
 A: The problem is in the picture.  In the picture, there are five hexagons adjacent to each pentagon.  Moreover, each hexagon is adjacent to exactly two pentagons.  This gives that
$$
\frac{12\cdot5}{2}=30
$$
hexagons are needed, not $20$.  Therefore, the given figure is does not match the given data.
A: The issue is that the picture depicts not the conventional soccer ball (a truncated icosahedron) but rather something a little different, the chamfered dodecahedron, also known as a truncated rhombic triacontahedron. This actually does have 120 edges.

So, in a sense, you were right both times, but just thinking about different polyhedra!

It's interesting to note that these are both examples of Goldberg Polyhedra, polyhedra made from only pentagons and hexagons -- although the faces are not necessarily regular (and in the chamfered dodecahedron, they are not). 
A: What's wrong is that your diagram does not have the usual pattern of pentagons and hexagons that a soccer ball usually has. The seams of a soccer ball are given by projecting centrally the edges of the truncated icosahedron onto a sphere:

In particular, no vertex on a truncated icosahedron is shared by three hexagons, but that is not the case for the polyhedron in the diagram, which (per pjs36's answer) is called a chamfered dodecahedron. (A truncated icosahedron is also the shape of the molecular structure of buckminsterfullerene a.k.a. buckyballs.)
