Combinations for a 1x2x3

[BirdCubing]

Hi everyone.
Recently, I just got a 1x2x3 for Christmas and I was trying to think about this question. Can anyone help me try to calculate how much combinations a 1x2x3 have?

 

[Bingus cuber]

I think 48. I looked it up tho.

 

[Discomantis]

Hi everyone.
Recently, I just got a 1x2x3 for Christmas and I was trying to think about this question. Can anyone help me try to calculate how much combinations a 1x2x3 have?

I’ll take a stab at it:

The middle pieces I will call edge pieces and the other pieces I will call corner pieces

For each corner, you cannot twist it. For each edge however, you can flip it.

Here is what each turn will to do the pieces:

U2 + D2: 1 corner swap 0 edge swap 0 flips
E2: 1 edge swap
R2 + L2: 1 corner swap, 1 edge flip

Here is what we can figure out from this:

Corner parity could be even or odd because you can do a turn that does an odd swap. Edge parity is the same way.

We can also figure out that because we can do an odd swap of corners independently and we can do an odd swap of edges independently, we can have any combination of edges and corner swaps.

We can also figure out that we can have a single edge flip because you can do a single edge flip the one turn.

In simpler terms: We can swap any two edge pieces and any two corner pieces, and we can do a single flip of edges.


So, we can calculate that we will have 4! possibilities of corners (4 possibilities for the first corner, 3 for the next, 2 for the next, and 1 for the last one)

and we can calculate that we will have 2^2 * 2 possibilities for edges (2 possibilities for the first one as well as two possible ways to flip the edge, and one possibility for the second edge and two ways to flip the edge)

So you have 4! * ((2^2 * 2))

Or 192 possibilities

Please let me know if I am wrong

 

[Jorian Meeuse]

I’ll take a stab at it:

The middle pieces I will call edge pieces and the other pieces I will call corner pieces

For each corner, you cannot twist it. For each edge however, you can flip it.

Here is what each turn will to do the pieces:

U2 + D2: 1 corner swap 0 edge swap 0 flips
E2: 1 edge swap
R2 + L2: 1 corner swap, 1 edge flip

Here is what we can figure out from this:

Corner parity could be even or odd because you can do a turn that does an odd swap. Edge parity is the same way.

We can also figure out that because we can do an odd swap of corners independently and we can do an odd swap of edges independently, we can have any combination of edges and corner swaps.

We can also figure out that we can have a single edge flip because you can do a single edge flip the one turn.

In simpler terms: We can swap any two edge pieces and any two corner pieces, and we can do a single flip of edges.


So, we can calculate that we will have 4! possibilities of corners (4 possibilities for the first corner, 3 for the next, 2 for the next, and 1 for the last one)

and we can calculate that we will have 2^2 * 2 possibilities for edges (2 possibilities for the first one as well as two possible ways to flip the edge, and one possibility for the second edge and two ways to flip the edge)

So you have 4! * ((2^2 * 2))

Or 192 possibilities

Please let me know if I am wrong

192 is in some sense correct, however, reorientations are a thing. 192 / 4 = 48 is the answer most consistent with other calculations like this (we also don't multiply 43 quintillion by 24).
A way to arrive here would simply be: orient the cuboid such that the left edge is solved. Now, the other edge has 2 possible states (tlipped or not flipped), and the corners have 4! states. Then, 4! * 2 = 48