How many configurations are possible in a Rubik's cube? A Rubik's cube has 54 visible facelets. There are 6 colours and each colour has 9 facelets. How many configurations are possible? How do you find that?
 A: Noting that there are indeed some restrictions on the number of edges (12) and corners (8) of the cube we have, according to Wikipedia, about 43 quintillion possibilities:
$$8!3^7\frac{12!}{2}2^{11}=43252003274489856000$$
The actual explanation as to why these numbers is on the Wikipedia link provided above:

There are 8! ways to arrange the corner cubes.  Each corner has three
  possible orientations, although only seven  (of eight) can be oriented
  independently;  the orientation of the eighth (final) corner depends
  on the preceding seven, giving 3^7 possibilities.  There are 12!/2
  ways to arrange the edges, restricted from 12! because edges must be
  in an even permutation exactly when the corners are. Eleven edges can
  be flipped independently, with the flip of the twelfth depending on
  the preceding ones, giving 2^11 possibilities.

Old (and wrong) answer
Using combinatorial numbers you have that the possible orderings correspond to the multiset permutation of the set:
$$S=\{C_1,...,C_1,C_2,...,C_2,...,C_9,...C_9\}$$
where $C_i$ denotes one element of colour $i$. This set has $9\cdot 6=54$ elements and 9 elements for each colour, then the number of possible ordenations follows:
$${54 \choose 9,9,9,9,9,9}=\frac{54!}{9!9!9!9!9!9!}\approx 1\cdot 10^{38}$$
