Paths on a cube 
Eight identical unit cubes are stacked to form a $2\times2\times2$ cube, as shown.  A "short path" from vertex $A$ to vertex $B$ is defined as one that consists of six one-unit moves either right, up or back along any of the six faces of the $2$-unit cube.  How many "short paths" are possible


I used the reasoning that a path needs to go right (R), up (U), and back (B) two times each, exactly. Thus, each path is going to be some combination of RRUUBB. There are $\frac{6!}{2!2!2!}=90$ combinations.
However, the answer turns out to be 54. Where did I go wrong in my reasoning?
 A: You have to move over the surface of the 2-unit cube; you're not allowed to go through the centre of the cube.
A: Try this diagram.  The red numbers are the number of short routes from A on the outside

A: The problem here is that you've assumed that you're moving from block to block, although you are only allowed to move from face to face. Moving from face to face gives an answer of $54$ as desired.
A: The first $3$ moves must be $2$ steps in one direction and $1$ step in another direction (for example Right Up Right). This is because if you go in one direction $3$ times in a row you find yourself no longer on the cube. If you go Right once, Left once, and Up once then you find yourself at the center of the cube.
There are $3$ directions in which you can go $2$ steps, $2$ remaining directions in which you can take $1$ step, and $3$ ways you can order these (RRU, RUR, URR) giving us $3\cdot2\cdot3=18$ ways we can successfully take our first $3$ steps.
There are only $3$ ways to go the remaining $3$ steps because the directions are already chosen by the first $3$ steps and there are $3$ ways to order the directions (RRU, RUR, URR).
Multiplying, this gives us $18\cdot3 = \color{red}{54}.$
