(1) Note that 4 of the 2-D tetrominoes (I, L, T, Z) each has a perimeter of 10, and O has perimeter 8. Use 10 as perimeter for every piece, an overestimate. (2) In the Tetris Lights toy, the assymetric pieces (L, Z) are duplicated, yielding 7 distinct pieces. The number of ways to order these is 7! = 5040. The assymetric pieces may appear either flipped or unflipped, for a factor of 2^4 = 16. (Reduce the exponent to disallow flipping, and/or to count mirror symmetric pairs of solutions as equivalent.) So there are 5040 * 16 = 80640 ways to choose the first through last piece. Note, the instructions suggest piece I lie on the tabletop, because the power cord attaches to it. We do not make that restriction. (3) Each piece can be rotated and translated, but not flipped; flipping is already accounted for. Pieces have 1 (O), 2 (I, Z), or 4 (L, T) rotational positions. For an upper bound, suppose the first piece has 4. In particular, the first piece can be in any of its 4 rotational positions, for a factor of 4. (4) The first piece rests on the tabletop, in an arbitrary place (left-right translation is irrelevant). At least one of its 10 units of perimeter touches the table and is thus unavailable for touching another piece. Its "available perimeter" is 10 - 1 = 9. (5) Place the second piece. For each of the units in the current available perimeter (now 9), and each of the units in the perimeter of the second piece (upper bound is 10), rotate and translate the second piece so those two units contact. Thus, there are 9 * 10 = 90 ways to place the second piece. Note, many of these 90 ways will be equivalent to others, or will be illegal because the new piece would interpenetrate previous pieces or the table. That is OK, because we seek only an upper bound. (6) Bound the available perimeter of the current construction (first and second pieces). Prior active perimeter (9) + perimeter of new piece (10) - 2 units used up in contact = 17. (7) Place the third piece. There is an upper bound of 17 (current active perimeter) * 10 (perimeter of new piece) = 170 ways to place the third piece. After placing it, the new active perimeter has an upper bound of 17 + 10 - 2 = 25. Note, many configurations will be created more than once, because placing the pieces (e.g., second and third) in the opposite order may generate the same result. This is OK, because we seek only an upper bound. The claim is that this process will generate every solution at least once. (8) Place the fourth piece. 25 * 10 = 250 ways. New active perimeter is 25 + 10 - 2 = 33. (9) Place the fifth piece. 33 * 10 = 330 ways. New active perimeter is 33 + 10 - 2 = 41. (10) Place the sixth piece. 41 * 10 = 410 ways. New active perimeter is 41 + 10 - 2 = 49. (11) Place the seventh (last) piece. 49 * 10 = 490 ways. New active perimeter is not of interest. (12) Combine above factors. 80640 * 4 * 90 * 170 * 250 * 330 * 410 * 490 = 8064 * 4 * 9 * 17 * 25 * 33 * 41 * 49 * 10^7 = 8064 * 9 * 17 * 33 * 41 * 49 * 10^9 = 81796708224 * 10^9 < 10^20