Twenty-seven identical white cubes are assembled into a single cube, and the outside of the cube is painted in the following manner.
- The top face, the bottom face and, one of the remaining faces is painted black.
- The remaining three faces are painted blue.
The cube is then disassembled and rebuilt randomly.
Determine the probability that the faces earlier painted black once again reappear as black and the faces painted blue are once again blue.
There are 3 pieces that have black paint on one face and are otherwice unpainted; these have to go back to ones colored in that manner and in the correct orientation:
3/27 * 2/26 * 1/25 * (1/6)^3
And the same for the three one-painted-blue faces:
3/24 * 2/23 * 1/22 * (1/6)^3
There are 12 edge pieces, each with one blue face and one black face:
12/21*11/20*10/19*9/18*8/17*7/16*6/15*5/14*4/13*3/12*2/11*1/10 * (1/24)^12
as this time there is only one correct orientation out of 24 for each cubelet.
There are 4 corner pieces with 2 black and one blue face:
4/9 * 3/8 * 2/7 * 1/6 * (1/24)^4
Similarly for 4 pieces with one black and two blue:
4/5 * 3/4 * 2/3 * 1/2 * (1/24)^4
The wholly unpainted cubelet will certainly go in the remaining place.
The product of these products is
80011342046416036521274602280229855382715025
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16451470922752763131039439848168064822029056637423428972227953375407712550837303436858359808
or about
4.8634764892517 x 10^-48
This is to exactly replicate the coloration of the original, including its orientation. If it's allowed to be in any orientation, there are 6*2 = 12 times as many solutions, as:
The blue face that's between two other blue faces can be in any of 6 places. The cube can be rotated about the axis perpendicular to that face, but of the four rotations two pairs are indistinguishable, so there are only 2 apparent rotations.
12*4.8634764892517 x 10^-48 = 5.83617178710204 x 10^-47 for the greater probability.
Edited on April 13, 2023, 10:17 am
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Posted by Charlie
on 2023-04-13 10:16:24 |
solution (if I did it right)
