permutation counts

how many ways can you scramble a puzzle

2x2x2

there are 8 pieces on the 2x2x2; any particular one can be taken as having a fixed location and orientation, as there is nothing identifying the orientation of the entire cube

out of the 7 remaining pieces, they can be permuted in any way; however, the orientation of the final piece depends on the orientation of the other 6

and there are 3 ways to orient an individual piece, so multiplying orientation and permutation together gives

$$3^6 \times 7! = 3\,674\,160$$

3x3x3

there are 8 corner pieces, 12 edge pieces, and 6 center pieces on the 3x3x3; the center pieces do not move relative to each other and can thus be taken as having a fixed location to identify the orientation of the entire cube

the edge pieces have 2 orientations each; the orientation of the final edge depends on the orientation of the other 11 edges

the corner pieces have 3 orientations each; the orientation of the final corner depends on the orientation of the other 7 corners

the edge pieces can appear in the location of any other edge piece and the corner pieces can appear in the location of any other corner piece

however, the permutation of the whole puzzle must be an even permutation; the total number of two-piece swaps must be an even number, reducing the number of permuations by a factor of 2

so multiplying edge orientation, corner orientation, edge permutation, and corner permutation together gives

$$2^{11} \times 3^7 \times (12! \times 8!) \div 2 = 43\,252\,003\,274\,489\,856\,000$$