Define a move to be a $90$ degree turn of $1$ of the $6$ sides of the cube clock-wise or counter-clockwise.
So for a $3\times 3\times 3$ cube I know that the number of possible unique states are $8!⋅12!⋅2^{10}\cdot 3^7.$ I have two questions:
1) Say I start with a solved cube, using only $1$ move, how many different states can be achieved? What in about $2$ moves?
2) The number of states that can be achieved in maximum $18$ moves is less than the total number of moves. How is this proven?
So for the first question, in $1$ move, each side can be turned in $2$ ways, so we get $2$ distinct states per side, which is a total of $2\cdot 6=12$ states.
Same reasoning with $2$ states but here we have to be careful because twisting one side twice clockwise gives the same state as twisting that side twice counter clockwise. Also we don't nessecarily need to move $1$ side twice in a row. This is were I get stuck.
Questions:
- Is what I've done for the first question regarding only 1 move correct?
- In order to solve question 2) I assume I need a general expression for the number of unique states one can obtain with $n$ moves. But my brain seems to protest here. Any suggestions on a good start/strategy?