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Two questions about path problem:

  1. We have a $2*2*2$ Cube. An ant wants to go from $A$ to $B$ and it can only walk on the lines . The path must be the shortest. (so 2 Up , 2 Left , 2 Forward -or backward- depending on how you name the directions) How many ways it can do it? enter image description here

  2. Now we Have $3*3*3$ with conditions of question 1, How many ways the ant can go from $A$ to $B$?

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My problem is the book which I found the problem in it solved the second one with this formula:

$\frac{9!}{3!3!3!} = \binom{9}{3} * \binom{6}{3} * \binom{3}{3} = 1680$

But if you solve the first one with this formula: $\frac{6!}{2!2!2!} = \binom{6}{2} * \binom{4}{2} * \binom{2}{2}$ you'll get 90 and it says the answer is wrong and in the solution, it gives the first 3 vertices number 1 and then it sum them until it gets the answer 54. My question is which answer and solution is right? How to solve this questions?

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I am guessing that the ant must stay on the surface of the cube (it can't go to the interior); in which case, the solution to number 2 is incorrect.