>>1457008I think this logic is incorrect. For example, on the 3rd row, just because the 2 can be in the 4th position or the 5th position doesn't mean there's only two possibilities. There's more. For example, 54123, 45123, 34125, 43125, etc. There's two possibilities for the position of the 2, but more possibilities for the rearranging of other numbers you're not counting.
My immediate instinct is that the answer is 60. There's 120 total number combinations, and since 1 and 2 appear equally in all positions, 1 will be in front of 2 half the time, and 2 will be in front of 1 half the time. Therefore half of them will be invalid.