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.
  C(14,8) × 8! × 6⁸ = 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.
  C(14,7) × C(2, 1) × 8! × 6⁷ × 3¹ = 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.
  C(14,6) × C(2, 2) × 8! × 6⁶ × 3² = 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.

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