Posted by pork (68.55.244.39) on March 26, 2003 at 21:04:47:

In Reply to: What is the answer to the end of your story? posted by Flower Pagan (68.35.112.137) on March 26, 2003 at 15:24:29:

FP,

For all I know, Bart has probably known the answer for years. Math is his forte, after all, and the answer to the "Rubik's Cube" problem is actually an old bit of trivia. Sometimes I wish I were better at the kind of visual imagination necessary for arithmetic proficiency; my meager talents afford but a glimpse of that facility.

Anyway, the answer... let's assume Melvin just rotated the cube, instead of dissembling it.

The forumla:

(8!*(3^7)*12!*(2^10)) = approximately 4.3e+19

You can arrange the corner cubes 8!*(3^7) ways - 8 corners, 3 orientations each; given 7 corner-cube placements, the last corner is constrained to 1 of the 3 orientations.

You can arrange the edge cubes 12!*(2^11) ways - 12 edges, 2 orientations each; given 11 arrangements, the last edge is constrained to 1 of the 2 orientations.

This covers all combinations; rearranging the center cubes would repeat patterns.

Also, since 3x3x3 is a cube has an odd number of cubes along each length (as opposed to the 4x4x4 Rubik's Revenge, which has an even number), you cannot switch 2 cubes without affecting the other cubes, thus you cannot make half of the above-calculated combinations. To account for this, the formula above replaces 2^11 with 2^10.

Some aspects of the solution are immediately apparent. Most people have to actually experiment with the cube to discover the constraints that are not apparent, however, which would take a while. It's much easier to calculate the number of combinations made possible by dissembling the cube: 8!*(3^8)*12!*(2^12), which increases the figure by 12 times.

^(oo)^