How can this technique be applied to a different problem? Here is the problem (copy and pasted if you don't want to click on the link).

Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex. Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal probability. What is the probability that no two ants arrive at the same vertex?
Source: 2005 AMC 12B, problem 25

The answer is $\dfrac5{256}$, but I was wondering if the same techniques used in this problem could be used in a similar one with a cube ($8$ ants instead of $6$).
If so, could someone show me how? Thanks!
Note: Unsure about tags as this has to do with events with states.
 A: Color four vertices of the cube together with  the ants sitting there blue, and   color the other four vertices together with  the ants sitting there red, as shown in the following figure:

Let $V_b$ be the set of blue vertices. Each blue ant marches from its  vertex to one of the red vertices, but not to the red vertex opposite to the starting vertex. Encode the combined displacement of the blue ants as follows: For each ant do not take down the number of its red target vertex, but the number of the blue vertex opposite to that red target vertex. In this way the combined displacement of the blue ants determines a fixed point free map $f:\>V_b\to V_b$.  When the combined displacement is successful this map is a fixed point free permutation of $V_b$. Conversely: Any such permutation defines a successful displacement of the blue ants. As $|V_b|=4$ such a permutation is either cyclic (there are $6$ of these) or consists of two transpositions (there are $3$ of these); whence there are only $9$ successful choices for the blue ants. Since there are $3^4$ choices in all, the probability that the blue ants are successful comes to ${9\over 3^4}={1\over9}$.
The same reasoning applies to the red ants, and since their success is independent of what the blue ants achieve it follows that the probability $p$ of a successful outcome for both groups simultaneously is given by$$p={1\over9}\cdot{1\over9}={1\over81}\doteq 1.235\>\%\ .\ .$$
