Proving a graph has no bridges Having a bit of trouble with this question:
Suppose G is a 4-regular connected graph with a planar embedding such that every face has degree 3 or 4, and further that any 2 adjacent faces have different degrees.
For the first part, I have to prove that G has no bridges and hence that every edge in G is on the boundary of 2 distinct faces. 
For the second part, I must determine precisely the number of vertices, edges, faces of degree 3, and faces of degree 4 in G.
Something tells me to use Euler's formula for the second part but I'm quite confused at the first part. 
Any help would be greatly appreciated!

For the first part of the answer see 100 roads in a city, 1 is closed. Here you can substitute 100 by every even number, such as 4.
For the second part, let $e$ be the number of edges, $f_4$ the number of quadrilaterals, $f_3$ be the number of triangles and $v$ the number of vertices. Notice that by the regularity you get the number of edges in terms of vertices. In particular, you have  $$4v=2e. $$ By a similar double counting argument: $$3 f_3+4 f_4= 2e.$$  Then you can use Euler's formula, which says $$f_3+f_4-e+v=2.$$ 
You also know something more, every triangle is adjacent to three quadrilaterals and by summing up the quadrilaterals this way, you overcount by a factor of $4$. Hence,
$$3 f_3 = 4 f_4.$$
Now we solve the system with the four equations. This gives
$$\begin{align}
f_3 & =  8,\\ f_4 &= 6, \\  e &= 24, \\ v&=12.
\end{align}
$$
Here is an example:

This is the graph of the cubeoctahedron. Should be the only possibilty.
