Balance an octahedron on its point, so that the bottom vertex is the south pole and the top vertex is the north pole. The other four corners touch the equator at 90° intervals. Now push the edges out towards the sphere. The four horizontal edges become the equator and the vertical edges become lines of longitude that connect the equator with the north or south pole. A face on the octahedron becomes a section of the sphere, like a piece of orange peel.
Now that the solid is mapped onto the sphere, we can apply Euler's formula, which says v-e+f = 2. You can see the proof here. Revisit the five regular solids and you'll see that vertices - edges + faces always equals 2. We can use this relationship to count the faces of a solid before it is actually constructed.
Pick an example out of thin air. Let each corner consist of a pentagon and two squares. This adds up to less then 360°, so we're ok.
Let v be the number of vertices. Each vertex has three edges coming into it, and each edge is counted twice (two end points), so the number of edges is 3v/2.
Turning to faces, each vertex consumes one fifth of a pentagon and two quarters of a square. Thus f = v/5 + 2v/4 = 7v/10. Now apply Euler's formula and get v - 3v/2 + 7v/10 = 2, thus v = 10. There are ten corners, two pentagons, and 5 squares. The shape has a pentagon floor and ceiling, and five square walls.
We can count the faces edges and corners of any shape, as long as we know what each corner looks like.