In the case of four distinct digits, after you choose the four digits (in $\binom94$ ways), you must choose which of the four is to be repeated; since you can do this in $4$ ways, the actual number of unordered hands in this case is $4\binom94$, not $\binom94$. The factor of $\frac{5!}2$ to account for the order of the cards is fine.
Similarly, in the third case there are $\binom93$ sets of three distinct digits, but this doesn’t tell you which digit is the singleton; there are $3$ ways to choose it, so the actual number of unordered hands in this case is $3\binom93$. Here again your accounting for order is just fine.
Exactly four different on 6 means that there are some repetitions. Maybe one digit repeated 3 times or maybe 2 repeated two times, example: 111235, 224456. So you want to know these two type of variations and sum them.
There are named variations with repetition and you calculate them in two steps:
1) Think the different digits as groups so a number as AABCDA have 4 different groups: A, B, C and D. Over these 4 groups calculate the number of permutations with repetition with the multinomial coefficient
$$PR_{n}^{k_1,k_2,...,k_n}=\binom{n}{k_1,k_2,...,k_n}=\frac{n!}{k_1!k_2!\cdots k_n!};\ n=k_1+k_2+...+k_n$$
In you case you have two different setups as said above so
$$PR_1=\binom{6}{2,2,1,1}=\frac{2\cdot3\cdot4\cdot5\cdot6}{2\cdot2}=180\\
PR_2=\binom{6}{3,1,1,1}=\frac{2\cdot3\cdot4\cdot5\cdot6}{2\cdot3}=120$$
2) Now for each type of permutation you must calculate the number of variations because A, B, C or D could be any digit between 0 and 9. There are variations of 10 elements over 4 positions, so
$$V_n^k=(n)_k=n(n-1)(n-2)\cdots(n-k+1)\\
V=(10)_4=10\cdot9\cdot8\cdot7=5040$$
3) The total of different number for each setup will be $\sum_{i}PR_i\cdot V$ but we need to discount the combinations that start with zero(s) that are a 10% of the total because exist the same probability for any number in any position and they are 10 different numbers, so the probability that the first digit will be a zero is $\frac{1}{10}$.
And the amount $PR_1$ must be divided by two because the two groups of two elements will repeat the same configuration twice, i.e., we have (by example) the configuration of permutations AABBCD where A, B, C and D take values from 0 to 9 (that is counted on the variations) but there is a symmetric permutation as BBAACD that can repeat the same numbers. The same happen for C and D so we must divide again by two to fix the duplicities. These numbers are the permutations of each type of indistinguishable multiplicity, i.e, $2!=2$.
In a similar way happen with the second setup $PR_2$ with B, C and D. So we need to divide between the number of permutations of indistinguishable multiplicities that is $3!=6$ (and $1!=1$).
So the total of cases will be
$$C=\left(PR_1\frac{1}{4}+PR_2\frac{1}{6}\right)V\cdot\frac{9}{10}=(45+20)81\cdot 56=294840$$
Best Answer
This approach is quite hard because of "4 less or 3 less depending upon whether we selected 0 or not" part.
Try to start with the first digit instead. How many choices have you there? How many choices left for the next one? And so on...