Instead of dividing into cases, we use the Principle of Inclusion/Exclusion.
There are $4^6$ strings of length $6$ over our $4$-letter alphabet. Now we count the bad strings, in which one or more digits are missing.
There are $3^6$ strings with the digit $1$ mising, and also $3^6$ with $2$ missing, with $3$ missing, with $4$ missing.
So our first estimate for the number of bad strings is $4\cdot 3^6$.
However, when we added, we multiply counted the bad strings that have more than one missing digit. For example, there are $2^6$ strings with $1$ and $2$ missing, and the same for all $6$ ways to choose $2$ digits from $4$.
So our next estimate for the number of bad strings is $4\cdot 3^6-6\cdot 2^6$.
However, we have subtracted too much, for we have subtracted one too many times the $4$ strings with $3$ digits missing. So the number of bads is $4\cdot 3^6-6\cdot 2^6+4$.
More neatly, we can write the number of good strings as
$$\binom{4}{0}4^6-\binom{4}{1}3^6+\binom{4}{2}2^6-\binom{4}{3}1^6.$$
The method readily generalizes.
Your cases approach will work. For no repetitions, the number is clearly $9\cdot 8\cdot 7\cdot 6\cdot 5$.
For a single repetition, the repeated digit can be chosen in $9$ ways. For each way, its locations can be chosen in $\binom{5}{2}$ ways, and for every such way the empty spots can be filled in $8\cdot 7\cdot 6$ ways.
Double repetition is a little trickier. The two fortunate digits can be chosen in $\binom{9}{2}$ ways. For each such way, the locations of the larger digit can be chosen in $\binom{5}{2}$ ways, and then the locations of the smaller one can be chosen in $\binom{3}{2}$ ways. The remaining empty spot can be filled in $7$ ways.
Remark: We can alternately count the complement. This avoids the trickiness of the double repetition count, where it is all too easy to overcount by a factor of $2$. There are $9$ sequences with all entries the same. For $4$ the same and $1$ different, we have $9\cdot \binom{5}{4}\cdot 8$ choices. For $3$ the same and $2$ different, we have $9\cdot \binom{5}{3}\cdot 8\cdot 7$. And finally for $3$ the same and $2$ the same we have $9\cdot \binom{5}{3}\cdot 8$.
Best Answer
We should form two cases, one of 0 (because 0 can't be the first one) and one with all numbers 1-9, to find out how many numbers there are that use a digit 3 or more times. Then, we can subtract this from 90,000 (the number of five digit numbers). This will give us the desired number.
Case 1: 0 shows up more than 3 times.
When you think about it, the only way 0 shows up more than three times is if the number is 10,000; 20,000; 30,000; etc., up to 90,000. So we just have to add nine to the following case.
Case 2: Some number 1-9 shows up more than 3 times.
This is slightly more tricky. We start with 1, just to make things easy. But when you really start to think about it, 1's would have to fill 4/5 slots in the number for it to break the rules. Therefore, the numbers (if the 5th number is 2) would be as follows:
11112
11121
11211
12111
21111
and we can do the same thing for every other number, 0 and 2-9. However, 01111 does not count, so we have to make sure to get rid of that one as well. Therefore, there are 8 times 5, or 40 ways this can be done for the number 1 for 2-9, and 4 ways this can be done with 0. Overall, the number 1 can be made in 44 different ways. This is also true for 2, 3, 4, 5, 6, 7, 8, and 9 (try plugging them in above where 1 is now). This gives us 9*44 + 9 ways to break the rules, or 9*45 ways. This comes out to 405 ways.
Subtract this from 90,000 and you will see that there are 89,595 ways to accomplish your task.