# How many permutations of a Rubik's cube, with three fixed (adjacente) faces?

So suppose I have a Rubik's, but I can only rotate the faces Top, Left and Front. How many configurations are there in this cube?

In a normal cube, the 'face' (i.e. center) tiles are always fixed. In my particular cube, on top of that, three edges and one corner are fixed (i.e. the corner opposed to the movable faces and it's three adjacent edges).

In a regular cube, we know that

1. I can reach all permutations of the 8 corners.
2. I can rotate independently 7 of the corners.
3. I can reach all EVEN permutations of the 12 edges.
4. I can rotate independently 11 of the edges.

So, for a normal cube, the number of permutations that keep a given corner and its adjacent edges fixed is:

$$N_1 = 7! \times 3^6 \times \frac{9!}{2} \times 2^8$$

I think the situation with our special cube is similar. With the limited number of moves, I have four hypothesis:

1. I can reach all permutations of the 7 movable corners.
2. I can rotate independently 6 of the movable corners.
3. I can reach all EVEN permutations of the 9 movable edges.
4. I can rotate independently 8 of the movable edges.

If the hypothesis are correct, the number of permutations for my special cube is also:

$$N_2 = 7! \times 3^6 \times \frac{9!}{2} \times 2^8$$

But I'm not sure. The two cases are not trivially equivalent, and I can't show that with the limited moves, I still get all that freedom that I want.

So, how many permutations cen I get?

• ... do you have a special kind of cube with 9 corners? The ones in my world have only 8 corners. – Henning Makholm Nov 22 '17 at 15:31
• Oh boy! Thanks, typo fixed – user2506946 Nov 22 '17 at 15:31

A mindless way of getting the answer is to ask GAP, by adapting the example in the manual (https://www.gap-system.org/Doc/Examples/rubik.html):

gap> cubeULF := Group(
> ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19),
> ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35),
> (17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11) );
<permutation group with 3 generators>

gap> Size(cubeULF);
170659735142400

gap> Collected( Factors( last ) );
[ [ 2, 18 ], [ 3, 12 ], [ 5, 2 ], [ 7, 2 ] ]


So the number of elements in the subgroup generated by the moves U, L and F is $$2^{18} 3^{12} 5^2 7^2 = 170659735142400 ,$$ which agrees with your proposed number $N_2$.

• +1, though one might reasonably hope for something less mindless. – Henning Makholm Nov 22 '17 at 15:49