The number 3211000 is 7-special Define a positive integer $k$ to be $n$-special if it satisfies the following properties:


*

*It has $n$ digits (0, 1, ..., 9)

*The 1st digit is equal to the number of 0's in the decimal representation of $k$, the second digit is equal to the number of 1's, the third digit is equal to the number of 2's, etc


For instance, the number $3211000$ is $7$-special (since there are three 0's, two 1's, one 2, and one 3).
Questions: 


*

*How many $7$-special numbers are there? Can you prove it?

*For what positive integer $n$ does there not exist any $n$-special numbers?

*Is there an efficient algorithm to compute all $n$-special numbers, given any positive integer $n$?
Have fun and enjoy! :)
 A: Just for kicks, I googled the 10-digit example (6210001000) that I mentioned in the comment, and found that these n-special numbers are apparently called self-descriptive numbers.
For your part 1), it doesn't list any other examples for 7, and judging from the OEIS list, there aren't any others. (As a heads-up, there's some base conversion going on in the list; your example is listed as 389305 (which gives 3211000 in base 7))
(Addendum: This list gives the numbers as you actually want them, though there are a few less entries)
For part 2), the article lists the three values of $n$ without any (2,3,6, though I guess 1 doesn't either if you want to count it) 
For part 3), the article gives a general formula for getting more examples in higher bases, but only one per value of $n$. If I find anything about generating all (either more formulas or a proof that this is all of them), I'll update the answer. The formula itself is
$$(n-4)n^{n-1} + 2n^{n-2} + n^{n-3} + n^3$$
A: Here's a proof there are no other $7$-special numbers.
In a $7$-special number $a_0a_1a_2a_3a_4a_5a_6$, each $a_k$ counts the number of numerals that appear $k$ times among the digits.  Since there are $7$ digits in all, we have
$$0a_0+1a_1+2a_2+3a_3+4a_4+5a_5+6a_6=7$$
This immediately implies $a_4+a_5+a_6\le1$, so at least two of those digits are $0$'s and at most one is a $1$.  Now if any of them are equal to $1$, then one of the other digits is at least $4$.  To satisfy the displayed equation, that digit can only be the $a_0$.   We also have $a_1\ge1$.  But we can't have $a_1=1$, since we already have at least one $1$, so we must have $a_1\ge2$.  This means there are at least two $1$'s among the digits, the next one being either $a_2$ or $a_3$.  But this leaves at most three digits that can possibly be $0$'s, which contradicts the conclusion $a_0\ge4$.
So we now know that $a_4=a_5=a_6=0$.  This means there are at least three $0$'s among the digits, hence $a_0\ge3$.  On the other hand, it also means that there are no $4$'s, $5$'s, or $6$'s among the digits, so $a_0$ cannot be any of those numerals.  Hence we must have $a_0=3$.  This implies that $a_1$, $a_2$, and $a_3$ are all at least $1$.  Writing them as $a_k=1+b_k$ and plugging into the displayed equation (along with the established values $a_0=3$ and $a_4=a_5=a_6=0$), we find
$$b_1+2b_2+3b_3=1$$
This is clearly enough to conclude that $3211000$ is the only $7$-special number.
There are plenty of other ways to let the logic of this argument run.  I'd be quite happy to see something shorter and more to the point.
A: The general problem of an $n$-special sequence, i.e., a sequence $b_0$, ..., $b_{n-1}$ with the property that each number $i\,$ in the range $[0:n-1]$ occurs exactly $b_i$ times in the sequence, can also easily be solved. The easiest way known to me to get all $n$-special sequences with proof is the following.
Let $b_0$, ..., $b_{n-1}$ be an $n$-special sequence. Each $b_i$ is in the range $[0,n]$, because it refers to a count out of $n$ elements.  $b_i=n$ is impossible; all elements would be $i$, also $b_i$ , but we know $i<n$.
Hence each element $b_i$ contributes $1$ to the sum $\sum_{i=0}^{n-1}{b_i}$, i.e., this sum is $n$. More strictly, the sum of the positive $b_i$ is $n$. There are $n-b_0$ positive $b_i$. 
$b_0=0$ is also impossible (this also means no element is $0$, a contradiction). Thus there are $n-b_0-1$ positive $b_i$ with $i>0$, and their sum is $n-b_0$. Hence one is $2$ and the others (if existing) are $1$. We obtain
a) If $b_0\le 2$, then $b_i=0$ for $i\ge 3$, $n=b_0+b_1+b_2$.
b) If $b_0\ge 3$, then $b_{b_0}=1$, $b_i=0$ for $i\ge 3$ and $i\ne b_0$, $n=b_0+b_1+b_2+b_{b_0}$ .
c) If $b_0=2$, then $b_2=2$.
d) If $b_0\ne 2$, then $b_2=1$.
We now make a case distinction by $b_0$.
$b_0=1$: We have $b_2=1$, $b_i=0$ for $i\ge 3$, $b_1=2$, $n=1+2+1=4$, and $b_3=0$. $\quad$ $\quad$ $\ $ This is the solution 1 2 1 0.
$b_0=2$: We have $b_2=2$, $b_i=0$ for $i\ge 3$, thus $b_1\lt2$.
$\quad$ $\quad$ $\ $ If $b_1=0$, then $n=2+0+2=4$, $b_3=0$. This is the solution 2 0 2 0.
$\quad$ $\quad$ $\ $ If $b_1=1$, then $n=2+1+2=5$, $b_3=b_4=0$. This is the solution 2 1 2 0 0.
$b_0\ge 3$: We have $b_2=1$, $b_{b_0}=1$, $b_i=0$ for $i\ge 3$ and $i\ne b_0$, and $b_1=2$.
$\quad$ $\quad$ $\ $ Thus $n=b_0+b_1+b_2+b_{b_0}=b_0+2+1+1=b_0+4$, i. e., $n\ge 7$ and $b_0=n-4$.
$\quad$ $\quad$ $\ $ This is the solution (n-4) 2 1 (n-7 times 0) 1 0 0 0 for each $n\ge 7$.
