# How many configurations are possible in a Rubik's cube? [duplicate]

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?

## marked as duplicate by Hans Lundmark, Alex M., Davide Giraudo, Namaste, Stella BidermanJan 3 '17 at 12:47

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.

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:

$${54 \choose 9,9,9,9,9,9}=\frac{54!}{9!9!9!9!9!9!}\approx 1\cdot 10^{38}$$

• Some of the facelets are connected by an edge or corner. That has to be taken into account. – Casper Jan 3 '17 at 8:01
• @Casper true. I'll try to recalculate it taking these restrictions into account. – Jesús Ros Jan 3 '17 at 8:13