This only works with one face used for both the 6 and 9.
Both cubes need to have 0, 1 and 2. The remaining six digits can be
assigned to the cubes in \( \binom{6}{3} / 2 = 10 \) ways.
There are 5 choices for a number to go on the face opposite the
zero. Then put the lowest remaining number on an open face, and
there are 3 choices for a number to go opposite that one.
There are 2 choices for how to arrange the final two digits.
So now we are up to \( 10 \cdot 30^2 = 9000 \) unique ways to assign
digits to the faces.
Alex Yeilding said:
This only works with one face used for both the 6 and 9.
Both cubes need to have 0, 1 and 2. The remaining six digits can be assigned to the cubes in \( \binom{6}{3} / 2 = 10 \) ways.
There are 5 choices for a number to go on the face opposite the zero. Then put the lowest remaining number on an open face, and there are 3 choices for a number to go opposite that one.
There are 2 choices for how to arrange the final two digits.
So now we are up to \( 10 \cdot 30^2 = 9000 \) unique ways to assign digits to the faces.