In how many ways can you place the stickers on a Rubik's cube so that the cube is not solvable?

Earlier, I asked how many ways you can place the stickers on a Rubik's
cube so that the cube is not solvable.  I have an answer to that.
There are N=54!/(9!)^6 ways to place the 54 stickers on a cube (ignoring
permutations of like-colored stickers).  Most of those ways are in sets
of 24 ways that differ only in the position of the cube in space.
however, there are M=4 18!/(3!)^6 ways of placing the stickers so that
the cube is invariant under a 120-degree turn in space; these occur in
sets of 8 that differ only in the position of the cube in space.  These
are the only ways of placing the stickers that have less than 24-way
symmetry.  And there are R=6! 8! 12! 2^10 3^7 ways of placing the
stickers so that the cube is solvable; all of these occur in sets of
24.  So the number of ways of placing the stickers so that the cube is
not solvable is

(N-M)/24 + M/8 - R/24
 = N/24 + M/12 - R/24
 = 54!/(24 (9!)^6) + 18!/(3 (3!)^6) - 6! 8! 12! 2^7 3^6
 = 4212390092651019262132517598730500000 +
    45741696000 - 1297560098234695680000
 = 4212390092651017964572419409776516000

when cubes differing only by whole-cube moves are considered
equivalent.

As for the other ObPuzzles:

    How many of them have a piece with a pair of opposite colors
    adjacent?

    How many have duplicate pieces?

Dik Winter quite rightly points out that the much more fundamental
question is:

    How many have six differently-colored face centers?

without which the question of solvability is fairly silly.  Dik posits
that the answer is D=2080262517294685958156352617272800000 when cubes
differing only in whole-cube moves are counted separately; I haven't
verified his answer (made more different by the lack of explanation of
the derivation).  But assuming his number is correct, all of them will
occur in sets of 24, so there would be D/24 - R/24 =
86677604887278581589848025719700000 - 1297560098234695680000 =
86677604887277284029749791024020000 unsolvable ways to sticker the cube
with six different face centers, up to whole-cube moves.

-- Dan Hoey <Hoey@AIC.NRL.Navy.Mil>

It is 6! . 48!/(8!)^6 (i.e. number of ways to give the centers different
colors multiplied by the number of ways to color the remainder).  And
indeed, to get different cubes up-to whole-cube moves this must be
divided by 6!.

-- Dik T. Winter <dik@cwi.nl>

[ Editor: there are 24 whole-cube moves, not 6! = 720, and with six
different centers no stickering is fixed by any of them, so Hoey's D/24
counts these cubes up to whole-cube moves; D/6! = 48!/(8!)^6 is the
count with the colors of the six centers fixed. ]

[ Editor: Hoey's posting began with a dispute with Chris Pelley about
an earlier exchange in the thread "Rubik's Cube impossible states";
that part is omitted.  No answers to the questions about opposite
colors and duplicate pieces were collected. ]