Suppose 27 cubes are stacked together, suspended in the air, to form a larger cube.
The cube is then painted on all the exposed surfaces and dried.
The smaller cubes are then randomly permuted in both spatial position and spatial orientation to form another large cube.
What is the probability that new larger cube is identical to the original?
This is not homework, it was passed on by a friend who found it somewhere on the internet, and they do not remember where.
After painting, there are $4$ types of small cubes:
1) $8$ cubes with $3$ painted faces.
2) $12$ cubes with $2$ painted faces.
3) $6$ cubes with $1$ painted face.
4) $1$ completely unpainted cube.
Now, a cube of type $1)$ must be placed again on a vertex. This gives $8!$ possible arrangements. However, such cubes can be oriented in $3$ possible ways, so we also have $3^{8}$ orientations. This gives $8!\cdot 3^{8}$ possibilities for the vertices.
Similarly, for type $2)$ we obtain $12!\cdot 2^{12}$ possibilities (those cubes can be placed in $12$ positions, but can only be oriented in $2$ ways).
For type $3)$ we get $6!\cdot 4^{6}$ possibilities.
Finally type $4)$ has $24$ possibilities (it can only be placed in one position, but can be oriented in $24$ ways).
For the total number of possible reassemblings, note that every cube can be oriented in $24$ ways, and that there are $27!$ rearrangements. So we have $27!\cdot 24^{27}$ possibilities in all.
Hence the probability is $$\frac{8!\cdot 3^{8}\cdot 12!\cdot 2^{12}\cdot 6!\cdot 4^{6}\cdot 24}{27!\cdot 24^{27}}.$$