Distinct Geometric Patterns of a Rubik Cube

[naus]

Hello, I'm new to this forum. I was hoping someone could help me out with a problem of discrete geometric patterns. After disassembling a Rubik Cube, I thought of this problem:

You have a standard 3x3x3 Rubik Cube, and want to figure out how many distinct ways n blocks can be removed from the cube. I'm calling the center piece one of the blocks making up the 27 block cube. Thus, any of the cubes can be removed, including the center. My strategy was going to be to use successive differences to start figuring out a formula, but then I realized that I can't figure out how many distinct patterns there are for n=2, even. (Are the first two values of the sequence, for n=1 and n=2, respectively, 4 and 26?) I'm not terribly familiar with geometry - is there a simple way to go about this problem?

Thanks for any help.

 

[naus]

wow! I forgot the most important part! The colors don't matter. That is, I'm treating all blocks as the same, like they're all the same color.

 

[galactus]

I would think the number of ways to remove n cubes from the 27 would be the same as choosing n from 27.

C(27,n). Say, for instance, we want to know how many ways we can remove 3 from the 27.

C(27,3)=2925 and so on

To remove 10, C(27,10)=8,436,285

BTW, in case you are not familiar with the notation, \(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

! denotes a factorial. Do you know what that is?.

 

[naus]

I would think the number of ways to remove n cubes from the 27 would be the same as choosing n from 27.

C(27,n). Say, for instance, we want to know how many ways we can remove 3 from the 27.

C(27,3)=2925 and so on

To remove 10, C(27,10)=8,436,285

BTW, in case you are not familiar with the notation, \(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

! denotes a factorial. Do you know what that is?.

That's combinations without repetition, but that doesn't account for duplications of the same pattern - e.g., if labeling the "front" of the cube 1-9 from left to right, top to bottom, (1,2) is the same pattern as (3,6), because a 90 degree anti-clockwise rotation of (3,6) results in the same view as (1,2). There are 351 combinations for 2 blocks, but only a fraction of them are distinct patterns.

Maybe I'm not using my terminology right. When I say distinct, I mean unique. Example: there are 4 ways to remove one block.
1 - center block of the cube
2 - corner block (in 1-9 face, block 1, 3, 6, or 9)
3 - middle edge block (in 1-9, block 2, 4, 6, or 8)
4 - middle face block (in 1-9, block 5).

I can't find a combinations/permutations formula that gives that number. I don't see a way to produce it through division from combinations, either. One solution would be to find the combinations without repetition and subtract duplicates from that. That's difficult for me, though, because the number of duplicates isn't the same for each two-block pattern. For example, using the 1-9 face again, there are 24 of the (1,2) and (2,3) patterns combined (number in a face * 3, when subtracting from 351), but only 6 for the patterns (1,9) and (3,7), because there are 8 ways to get the (1,2) and (2,3) patterns on a single face, but only 2 ways to get the (1,9) and (3,7) patterns on a face. Counting mistakes are very easy to make when you have to subtract duplicates of each pattern individually, and you also have to be sure you've found every pattern.

 

[naus]

I'm taking a new stab at this problem. My method that I described above was completely incorrect, as far as I can see. I also don't have the spatial perception to count all of the patterns the long way, by figuring them all out.

Here it is: can this problem be altered to become a 3D coloring problem? Is anyone here familiar with Polya's enumeration theorem or Burnside's Theorem? If instead of being removed the blocks in question are instead colored differently than all the other blocks, there may be a way to count the patterns based on an analytic breakdown of the rotational symmetries and such.

I don't know much about this subject, honestly.

 

[Denis]

Give Rubik a phone call....