Lemma 2:
I claim that on the even degree n x n x n cube the parity of every orbit is independent of all the other orbits - noting that the concept of parity does not exist for the center orbits of the cube.

Proof:
I admit that this proof is much longer than that which uses permutations and group theory, however this is the best way I know to prove this lemma. It is less efficient than using group theory, but it is the only way I know to do this, and as long as it works it still achieves my goal to prove my combination formula for the n x n x n cube.

Here I will be using the notation of inner slice moves. Let R denote turning the outer most right face once clockwise. Let r[1] denote turning the r slice just behind the outer R face once clockwise. Let r[n] denote turning the nth inner slice counting from the first inner right slice r[1].

First note that on the even degree n x n x n cube that there are (n-2)/2 edge piece orbits. Now we must examine the effects of every one of our possible moves, and how they affect the parities of all the orbits of the cube.

1) Let us start with the outer face turns.
a) Let's examine the effects of an outer most face single quarter turn. This will cycle the four corners on that face in 4 cycle. Breaking down a 4 cycle into transpositions we see that a 4 cycle is an odd permutation. Therefore we know that turning an outer face by one quarter turn will chance the parity of the corner orbit, from even to odd or vice versa. There are 4n-8 total edge pieces on an outer face, and these are spread evenly over the (n-2)/2 edge orbits. Therefore we know that there are 8 pieces from each edge orbit resting on the outer face.

Definition: The concept of leftness and rightness for an edge within a single orbit will be defined as follows. Given the orbit in which the piece r[k] k<=(n-2)/2 is contained, all the right positions of that orbit are as follows { UFr[k] , LUf[k] , BUl[k] , RUb[k] , FLd[k] , FRu[k] , BRd[k] , BLu[k] , FDr[k] , LDf[k] , DBl[k] , DRb[k] } and let the left positions be defined as the other 12 positions in that edge orbit.

Examining the outer face quarter turn, we can see that it will four cycle all of the (n-2)/2 left edge orbits, as well as all of the (n-2)/2 right orbits on that face. In each of the (n-2)/2 edge orbits the permutation created after an outer slice quarter turn is that of two 4 cycles of each pieces. This creates 6 transpositions necessary in each orbit to solve the edges back into their correct locations. These 6 transpositions perform an even permutation on all the edge oribts, thereby leaving the parity of all the edge orbits unaffected. Therefore an outer slice quarter turn will only change the parity of the corner orbit.

b) Now examine an outer slice double turn. An outer slice double turn is just two quarter turns performed on the same face in succession. Based on our previous statements, we see that the first quarter turn will change the corner parity from what it started from. Then the second quarter turn will change the parity back to what it started out. Since the parity can only be either even or odd, chaning it twice will in affect not change the parity. Also note that this doulbe turn will not affect the parity of any edge orbit. Therefore an outer slice double turn will not affect the parity of any orbit on the cube.

2) Now let us examine all inner slice quarter turns.
a) Let 1<=k<=(n-2)/2. Then turning any of the following slices will only affect one edge orbit: f[k] , r[k] , b[k] , l[k] , u[k] , d[k]. Pick any of these turns, which only affect the parity in which the UFr[k] piece is contained. Doing a quarter turn will not affect any corner at all. Therefore this turn can only affect the parity of the edge orbits. As we can see, in the orbit in which the UFr[k] piece is contained, doing a single inner slice quarter turn will 4 cycle the 4 edge pieces in that orbit that are contained on the r[k] slice. By 4-cycling the those pieces we have performed 3 transpositions, which is an odd permutation and will change the parity of that edge orbit only. This is because those 4 pieces in that orbit are the only edge pieces affected by this turn. Therefore doing any inner slice turn, represented by r[k], we only affect the orbit in which the UFr[k] piece is contained. Therefore This inner slice turn only affects that one orbit, independently of the others.

Examining an inner slice double turn, we see that it is really two quarter turns done on that slice in succession. By performing two quarter turns on that slice, we will change the parity of the affected edge orbit twice, effectively not changing the parity at all. Therefore an inner slice double turn will not affect the parity of any orbit on the cube.

Therefore we see that an outer face will only affect the parity of the corner orbit, and any inner slice turn r[k] will only affect the orbit in which the piece UFr[k] is contained, and leave the others unaffected. Therefore the lemma is proved, and the parity of every orbit is independent of all the others.

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