I take a [tex]n[/tex]-sequence to mean any permutation of the digits 0-9 of length [tex]n[/tex]. Furthermore, I'm assuming no number can be repeated.
If that's the case, then for any 4-sequence, the first digits place (leftmost) has 9 possible choices (1-9). Once that number is used up, we have 9 numbers left in the pool of digits, from which we select a 3-sequence. The number of possible 3-sequences would be [tex]\dfrac{9!}{3!}=60,480[/tex]. Multiply this by 9 (to account for the first digit's possibilities) and we get a grand total of [tex]\dfrac{9\times9!}{3!}=544,320[/tex] possible 4-sequences on [tex]\{0,\ldots,9\}[/tex].
