You Don't Need To Count

The number of edges in a cube is equal to the number of pairs of adjacent faces in a cube. Therefore, 12.

Okay, so firstly let us consider the number of ways we can choose a pair of faces in a cube; since we are choosing two elements out of six, then we have $C(6, 2) = \binom{6}{2} = 15$ ways of choosing any pair of faces on a cube.

Now, the only ways we can choose non-adjacent pairs are by taking two opposite faces; this can only be done in three different ways, since we only have six faces and only one face can be non-adjacent to another face; furthermore, the pair $(face \, 1,\, face \,2)$ is equivalent to $(face \, 2,\, face \,1)$ , and this is why we can have only three non-adjacent pairs of faces.

Thus, the total sought is $15 - 3 = 12$ pairs.

I was too lazy to even count the edges, so I just used Euler's formula. The cube, as we all know, has 6 faces and 8 vertices. Since $E = V+F-2$ ,

$E = 6+8-2 = \boxed{12}$

the most intuitive approach to solving this problem would be this... Since every face in a cube is adjacent to four other faces and there are in all six faces in a cube; so there are total 6x4=24 adjacent face pairs...but each of these pairs has been counted twice; once for each of the two faces participating in the pair. So the total number needs to be halved which gives you 12!!!

Just count and get your answer!........

As cube has six surfaces and in every surface there are actually two pairs in the up and bottom side consisting four edges (that is the front view from a single side !! ) So, the answer should be 6*2 = 12 pairs !!