Moving on the surface of a cube A $3 \times 3$ cube is composed of $27$, $1 \times 1$ cubes. Moving along the surface of the larger cube, how many ways are there to get from the closer top-left vertex, to the further bottom-right vertex?
Constraints:


*

*You may only move along the surface of the cube

*You may only move right, down, and forward

*You may only move on the edges of the smaller cubes

 A: Name the faces of the $3\times3$ cube touching the start A, B, and C. Name the faces touching the finish vertex D, E, and F. Any valid path lies within just two of these faces - one of A, B, and C, and an adjacent one of D, E, and F.
The number of paths that lie entirely within faces A and D is $9\choose3$. (Unfold the two faces into a $3\times6$ rectangle.) Similarly, there are $9\choose3$ paths that lie entirely within faces A and E, if A and E are adjacent, etc. There are 3 ways to choose one of A, B, and C and for each choice, two of the faces D, E, and F are adjacent, so this method counts $6{9\choose3}$ paths.
Unfortunately, this counts paths twice if they begin or end (but not both) by traversing an entire edge of the larger cube, and it counts paths three times if they both begin and end by traversing an entire edge of the larger cube.
The number of paths counted exactly twice is $3\cdot({6\choose3}-2)+({6\choose3}-2)\cdot3=108$, and the number of paths counted exactly three times is 6.
All told, then, there are $6{9\choose3} - 108 - 2\cdot6 = 384$ paths.
A: If you are interested in a general solution, for any $n x n$ cube, I will show you.
Odd $n$, where $n$ is the $k^{th}$ odd integer:
$\frac{(3n)!}{(n!)^{3}} - 3[2(\frac{(3n-3)!}{((n-1)!)^{3}}) + 6(\frac{(3n-4)!}{((n-1)!)^{2}(n-2)!}) + 6(\frac{(3n-5)!}{(n-1)!((n-2)!)^{2}}) + ... + (\frac{(n+1)!}{(k!)^{2}})(\frac{(n+1)!}{(n-1)!((n-k)!)^{2}})]$
Even $n$, where $n$ is the $k^{th}$ even integer:
$\frac{(3n)!}{(n!)^{3}} - 3[2(\frac{(3n-3)!}{((n-1)!)^{3}}) + 6(\frac{(3n-4)!}{((n-1)!)^{2}(n-2)!}) + 6(\frac{(3n-5)!}{(n-1)!((n-2)!)^{2}}) + ... + 2(\frac{(n+1)!}{(k)!(k+1)!})(\frac{(n+1)!}{(n-1)!(n-k)!(n-k-1)!})]$
A: At the beginning you can choose one of three options. Once you pick the first option you wont be able to move in one direction until you have done one of the two movements you can do at least three times.
Therefore wee need to count the nine digit numbers made of exactly three of each of the following digits: 1,2,3 such that there are not all three digits appear before the first third digit appears for example:
1,1,2,2,1,2,3,3,3 works but 1,2,3,1,2,3,1,2,3 does not because three of the same kind don't appear before all three types appear.
To do this count the correct numbers in which the third type of number appears in the fourth,sixth and seventh
A: Like Pascal's triangle, store 1 in the starting vertex, and for each vertex $a$ connected to a vertex with a stored value, add the values of all such connected vertices and store it in $a$.
Right, down, and forward are simply distance in the graph of the edges & points from the start, the further bottom-right vertex the furthest.
EDIT AGAIN: I count 512.
A: [Note: As @joriki points out, this is wrong, though perhaps elegant...]
Or much more easily, take the required 9 steps as follows, which will generate all possible paths once each.


*

*Choose one of the three initial directions. 

*For the next 7 steps, you will have exactly two directions to choose from.

*You will have no choice for the last move.


This gives $3\cdot 2^7$ paths.
