How to count - probability puzzle

[Quasar Chunawala]

A [imath]3 \times 3 \times 3[/imath] big cube consists of smaller [imath]1 \times 1 \times 1[/imath] smaller cubes. The big cube is painted black on the outside. Suppose we disassemble the cube and randomly put it back together. What is the probability of perfectly assembling the cube?

Solution attempt.

I thought about this puzzle today morning. There are:

[imath]8[/imath] corner(vertices) cubes with [imath]3[/imath] black faces - label them [imath]V[/imath]
[imath]12[/imath] edge cubes with [imath]2[/imath] black faces - label them [imath]E[/imath]
[imath]6[/imath] center of a face cubes with [imath]1[/imath] black - label them [imath]C[/imath]
[imath]1[/imath] center cube with no black face - label it [imath]O[/imath]

For the [imath]1 \times 1 \times 1[/imath] small cube, there are [imath]6[/imath] choices for the top face, and for each such choice, [imath]4[/imath] choices for an adjacent face. So, there are [imath]24[/imath] distinguishable orientations of the unit cube.

[imath](\forall V)[/imath] cubes, the probability it is correctly oriented = [imath]\frac{1}{24}[/imath]
[imath](\forall E)[/imath] cubes, the probability it is correctly oriented = [imath]\frac{1}{24}[/imath]
[imath](\forall C)[/imath] cubes, the probability it is correctly oriented = [imath]\frac{4}{24}[/imath]
[imath]O[/imath] cube, the probability it is correctly oriented = [imath]1[/imath]

Furthermore, we can swap one corner cube for another, or one edge cube for another and so forth, without affecting the outer appearance. So, there should be an adjustment of [imath]\frac{27!}{8!12!6!1!}[/imath].

Do you guys think, the progress so far is correct? If not, do you have a hint/clue (without revealing/giving away entire solution)?

 

[Quasar Chunawala]

Here are the ensuing comments on Math SE and I attach the worked-out solution.

How to count - probability puzzle