MHB In how many ways can a cube with 8 different vertices be oriented?

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A cube with 8 distinct vertices can be oriented in 24 different ways. By fixing one vertex at a specific point, there are 8 possible choices for that vertex. For each fixed vertex, there are 3 adjacent vertices to choose from, leading to a calculation of 8 multiplied by 3. Once two vertices are determined, the positions of the remaining vertices are automatically established. This confirms that there are indeed 24 unique orientations for the cube.
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My solution:

Imagine that we have 8 points in space which must be occupied by the vertices of the cube.
Let A be one of the points. Fix anyone of the vertices at A. This can be done in 8 ways.
We have 3 different choices for an adjacent point (say B) because each vertex of a cube is connected to 3 other vertices.
Once A and B (being adjacent points) are determined, the other points are automatically determined. (Is this true? If so, how would I prove it?)
So we have 8*3 = 24 orientations.

Is this correct?
 
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Yes, there are 24 cube rotations, or "physical symmetries", as they are called in this article.
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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