How many paths are there from point P to point Q if each step has to go closer to point Q. Each of the six faces of a solid cube is divided into four squares as indicated in the diagram. Starting from vertex P paths can be travelled to vertex Q along connected line segments. Each movement along a path must take one closer to Q. How many possible paths are there from P to Q?
I got 96 different routes. I found 48 in these three sides shown and then multiplied by 2 to count for the other sides. Not sure if I am correct. What do you guys think?
 A: Every path from $P$ to $Q$ consists of two steps forward, two steps down, and two steps to the left. The question becomes: In how many ways  can we permute the string "FFDDLL"? The answer is
$$
\frac{6!}{2! \cdot 2! \cdot 2!} = 90
$$
Edit: Thanks to LotharNarins. This doesn't account for the fact that you can't go through the center. This happens exactly when the first three letters are all distinct, i.e. a permutation of "FDL". Because you need 3 steps to get to the center, and if two of the first three letters are the same, you've already gone past it.
There are $3!$ ways to arrange the letters "FDL". For each of these there are $3!$ ways to arrange the remaining 3 letters, so the final answer is
$$
90-3!\cdot 3! = 54.
$$
A: There's an algorithm for answering problems like this. Label your initial point 1, since there's 1 path to reach the starting point from the starting point. As the inductive step, label any unlabeled vertex adjacent to a labelled vertex with the sum of all adjacent labelled vertices. Since your graph has (an upper bound of) $9 \cdot 6 = 54$ vertices, this may take a little time, but should not prove challenging.
