As Eric noted in a comment, you're not taking into account the fact that there are four different positions for the fourth digit. The correct calculation is
$$
10^4-4\cdot10\cdot10+3\cdot10=9630\;,
$$
where the last term corrects for the fact that the second term counts each string of four identical digits $4$ times. Alternatively, not counting them in the second term, you could also write
$$
10^4-4\cdot10\cdot9-10=9630\;.
$$
What you are looking for is the Stars and Bars method. This method is used to count things where order does not matter and repetition is allowed.
We are choosing $9$ digits from $\{1,2,3,4,5,6\}$
Let me explain how we can get to the answer.
For example, let's say we want to pick the combination $123456123$. This can be "arranged" as such:
$$••|••|••|•|•|•$$
We picked $2$ ones, $2$ twos, $2$ threes, and $1$ of the rest.
Let's look at another possibility: $111222456$
$$•••|•••| |•|•|•$$
We have three ones, three twos, and one $4$, $5$, and $6$.
Look at what these pictures are showing: we have $14$ symbols and we are choosing $5$ of them to be a bar. Or we could look at it as choosing $9$ of them to be a dot (star)
So the answer to our original question is $$\binom{14}{9} = \binom{14}{5} = 2002$$
Best Answer
Yes, there does exist such a way. First you select a digit d from {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Then you select a digit e from ({0, 1, 2, 3, 4, 5, 6, 7, 8, 9}-d). Then you select a digit f from (({0, 1, 2, 3, 4, 5, 6, 7, 8, 9}-d)-e). You first select 0 for d, then 1, and so on until you get to 7. And, you always select the least digit first for e and f also, with the additional condition that d < e < f. List out the first sequence, 012, 013, 014, 015, 016, 017, 018, 019. Then list all the other numbers beneath them with the condition that for all numbers e and f, and with d held constant, the digits for e and f follow the natural number sequence down the column. Partition each set of sequences by d. The column rule only applies within each partition. (this description might come as incomplete or could use some revision).
The list thus goes:
I'll clarify the last part here:
Perhaps better, say we try to do the same thing in base 4. The entire sequence goes
In base 5 the process yields:
Thus, in base 10 the sum of the first 8 triangular numbers gives us the number of such combinations: +(1, 3, 6, 10, 15, 21, 28, 36)=120. Similarly, it should logically follow that for x digit numbers in base z, where x < z, or x=z, there exist +[T$_1$, ..., T$_ (z-(x+1))$] such combinations, where T$_n$ indicates the nth triangular number.
I guess the underlying idea I've used here lies in following the natural number sequence across the rows, and down the columns for the digits e and f also.