Find the number of routes from leftmost vertex to rightmost vertex Suppose that one can move from one vertex to another if, and only if,
  the two vertices are connected by a unique common edge. The number of
  routes that one can take from the leftmost vertex L through 6 edges
  and 5 intermediate vertices to the rightmost vertex R is .......


This is my scholarship practice exam (pre-university level) and the only way I did it correctly is by counting but I feel it was a bit messy but it can be simplified by writing with alphabets such as (u for up, d for down, s for go straight). 

However, there could be other solutions such as combination or permutation, probably. The answer key provided is 12. Please advise what you think and how to solve this problem.
The annoying bit are the two triangles at the bottom and top. You can get the answer by considering paths that don’t and do go through those triangles. For the ones that don’t, there are $2\cdot \
2\cdot 1\cdot1\cdot 2\cdot 1=8$ paths. 

For the ones that pass through the triangles, convince yourself that it’s only possible to pass through exactly one and not both. For the bottom triangle there are two possible paths (differing only at the R end). For the top triangle there are two possible ways to get to it from L and then only 1 way to get to R. 

So the answer is 12
Try labeling the vertices and writing down the resulting adjacency matrix. Then raise it to the power of 6, and read the entry in the (i, j) position, where i and j are the labels of the left- and right- most vertices, respectively. Since the adjacency matrix is 14 x 14, you should probably use a computer. I recommend using the Wolfram alpha website. Although it wouldn't be too hard by hand, since A^6 = (A^2)^2 A^2.