#or some beautiful proof that leverages the question's proximity to combinatorics better than i did

[cathodic-clairvoyant]

Fun math/counting problem for everyone

Okay so I know this will only be interesting to like maybe one other person but I came across a fun counting problem I wanted to share. For a given number of digits, n, how many numbers have a unique sequence of digits up to permutation? (So to be explicit about what the question is asking, 15 and 51 would both count as the same digit sequence since the digits of 51 are just a permutation of the digits of 15. Also, leading 0s are also included in the number. For example in the four digit case, 15 would be written as 0015.)  

If you’re at all math inclined or like number puzzles I do genuinely encourage you to try to figure this question out yourself partially because it's fun but mainly because I’m curious if anyone else can come up with a different/better solution than I did. I'll put my solution under the break. It's half informal proof (making it more rigorous is an exercise for the reader or w/e) and half just like a guided solution. If you're familiar with manipulating sums and can read basic math notation (if you're unfamiliar with the notation when defining the function, don't worry about it, it really doesn't matter I just think it looks pretty) you should be able to follow (hopefully, unless i've done something terribly wrong).