Here is how the number of candidate solutions is figured.

__Step one.__ Install the square tiles. 4:0, 4:1 and 4:2 have four different rotations, but 4:3, 4:4 and 4:5 have only two.

- Without loss of generality, place tile 4:0 in any of the 6 square locations, rotated in any direction.
- Place 4:1 in any of the 5 remaining locations, rotating it any of 4 ways (5 × 4 = 20)
- Place 4:2 in any of the 4 remaining locations, rotating it any of 4 ways (4 × 4 = 16)
- Place 4:3 in any of the 3 remaining locations, rotating it either of 2 ways (3 × 2 = 6)
- Place 4:4 in any of the 2 remaining locations, rotating it either of 2 ways (2 × 2 = 4)
- Place 4:5 in the 1 remaining location, rotating it either of 2 ways (1 × 2 = 2)

__Step two.__ Install the hexagonal tiles. The 14 tiles labeled 6:0 through 6:D have 6 rotations (call them *6Rs*), but 6:E and 6:F have only 3 rotations (*3Rs*). This leads to three cases:

- If eight 6Rs are chosen…
- There are C(14, 8) (combinations) ways to select them.
- There are 8! (factorial) ways to assign them to locations on the polyhedron.
- Each of the 8 tiles can be rotated 6 ways.

^{8}= 3,003 × 40,320 × 1,679,616 = 203,369,517,711,360. - If seven 6Rs and one 3R are selected…
- There are C(14,7) ways to select the 6Rs, and C(2, 1) ways to select the 3R.
- There are 8! ways to assign the eight hexagons to locations.
- 7 tiles can be rotated 6 ways, and 1 tile 3 ways.

^{7}× 3^{1}= 3,432 × 2 × 40,320 × 279,936 × 3 = 232,422,305,955,840. - If six 6Rs and both 3Rs are used…
- There are C(14,6) ways to select the 6Rs and C(2,2) ways to select the 3Rs.
- There are 8! ways to assign the eight tiles to locations.
- 6 tiles can be rotated 6 ways, and 2 tiles 3 ways.

^{6}× 3^{2}= 3,003 × 1 × 40,320 × 46,656 × 9 = 50,842,379,427,840

__Step three.__ Multiply the number of square configurations by the number of hexagonal configuations: 15,360 × 486,634,203,095,040 = 7,474,701,359,539,814,400.