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 $9\cdot \binom{5}{3}\cdot 8$.
Let $N_7 = \text{what you are looking for}$
However, this consists of one set that doesn't use any zeros - $(NZ)_7$ - and another that does. The number in the set that uses zeros can be defined recursively as $6(NZ)_6$ because the zero can be placed in 6 different places.
Notice that $(NZ)_7 = \frac{\text{pair of two numbers that are divisible by 3}}{9c2} \cdot(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3)$
This is because the total sum of $1,2, ..8, 9$ is $45$ and we need to remove two numbers such that the sum of the others is still a multiple fo $3$ ie the two numbers we remove must be a multiple of $3$
Therefore, $(NZ)_7 = \frac{(NZ)_2}{9P2}(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3)$
Similarly, $(NZ)_6 = \frac{(NZ)_3}{9P3}(9\cdot8\cdot7\cdot6\cdot5\cdot4)$
$(NZ)_2$ can be counted, but also it can be expressed as $9\cdot3 - \frac{1}{9}\cdot 9 \cdot 3$. This is found because if the first is $0 \mod 3$, then there is an overcounting (1/3 of such cases are false $1/3 *1/3 = 1/9$)
Similarly, $(NZ)_2=9\cdot 8 \cdot 3 - \frac{2}{8} \cdot \frac{2}{3} \cdot 9 \cdot 8\cdot 3$. Here, the second part is found by considering the probability that the second number chosen is the same modulus as the first ( $2/8$), which results in only $1/3$ of the possible last digits (so we subtract out $2/3$)
Therefore,
$$N_7 = \frac{9 \cdot 3 - 3}{72}(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3) + 6 \cdot \frac{9\cdot8\cdot3-4\cdot9}{504}(9\cdot8\cdot7\cdot6\cdot5\cdot4) = 190080$$
This is definitely not my most eloquent answer, so please ask me questions if something is confusing, and I'll try to explain my thought process.
Best Answer
Let's confirm your solution.
There are $10^9$ strings of length $9$ that can be formed by using the ten decimal digits with repetition. From these, we must exclude those strings in which at least one odd digit is missing.
Let $A_i$ be the set of outcomes in which the digit $i$ is excluded, where $i \in \{1, 3, 5, 7, 9\}$.
Then, by the Inclusion Principle, the number of strings in which at least one odd digit is missing is $$\sum_{i} |A_i| - \sum_{i < j} |A_i \cap A_j| + \sum_{i < j < k} |A_i \cap A_j \cap A_k| - \sum_{i < j < k < l} |A_i \cap A_j \cap A_k \cap A_l| + \sum_{i < j < k < l < m} |A_i \cap A_j \cap A_k \cap A_l \cap A_m|$$
$|A_1|$: Since $1$ is excluded each of the nine positions in the string can be filled in nine ways. Hence, $|A_1| = 9^9$. By symmetry, $$|A_1| = |A_3| = |A_5| = |A_7| = |A_9|$$
$|A_1 \cap A_3|$: Since both $1$ and $3$ are excluded, each of the nine positions in the string can be filled in eight ways. Hence, $|A_1 \cap A_3| = 8^9$. By symmetry, $$|A_1 \cap A_3| = |A_1 \cap A_5| = |A_1 \cap A_7| = |A_1 \cap A_9| = |A_3 \cap A_5| = |A_3 \cap A_7| = |A_3 \cap A_9| = |A_5 \cap A_7| = |A_5 \cap A_9| = |A_7 \cap A_9|$$
$|A_1 \cap A_3 \cap A_5|$: Since $1$, $3$, and $5$ are all excluded, each of the nine positions in the string can be filled in seven ways. Hence, $|A_1 \cap A_3 \cap A_5| = 7^9$. By symmetry, $$|A_1 \cap A_3 \cap A_5| = |A_1 \cap A_3 \cap A_7| = |A_1 \cap A_3 \cap A_9| = |A_1 \cap A_5 \cap A_7| = |A_1 \cap A_5 \cap A_9| = |A_1 \cap A_7 \cap A_9| = |A_3 \cap A_5 \cap A_7| = |A_3 \cap A_5 \cap A_9| = |A_3 \cap A_7 \cap A_9| = |A_5 \cap A_7 \cap A_9|$$
$|A_1 \cap A_3 \cap A_5 \cap A_7|$: Since $1$, $3$, $5$, $7$, and $9$ are all excluded, each of the nine positions in the string can be filled in six ways. Hence, $|A_1 \cap A_3 \cap A_5 \cap A_7| = 6^9$. By symmetry, $$|A_1 \cap A_3 \cap A_5 \cap A_7| = |A_1 \cap A_3 \cap A_5 \cap A_9| = |A_1 \cap A_3 \cap A_7 \cap A_9| = |A_1 \cap A_5 \cap A_7 \cap A_9| = |A_3 \cap A_5 \cap A_7 \cap A_9|$$
$|A_1 \cap A_3 \cap A_5 \cap A_7 \cap A_9|$: Since $1$, $3$, $5$, $7$, and $9$ are all excluded, each of the nine positions in the string can be filled in five ways. Hence, $|A_1 \cap A_3 \cap A_5 \cap A_7 \cap A_9| = 5^9$.
Thus, by the Inclusion-Exclusion Principle, the number of strings of length $9$ in which at least one odd digit is missing is $$5 \cdot 9^9 - 10 \cdot 8^9 + 10 \cdot 7^9 - 5 \cdot 6^9 + 5^9$$ Therefore, the number of strings of length $9$ in which no odd digits are missing is \begin{align*} 10^9 - (5 \cdot 9^9 - 10 \cdot 8^9 + 10 \cdot 7^9 - 5 \cdot 6^9 + 5^9) & = 10^9 - 5 \cdot 9^9 + 10 \cdot 8^9 - 10 \cdot 7^9 + 5 \cdot 6^9 - 5^9\\ & = \binom{5}{0}10^9 - \binom{5}{1}9^9 + \binom{5}{2}8^9 - \binom{5}{3}7^9 + \binom{5}{4}6^9 - \binom{5}{5}5^9\\ & = \sum_{k = 0}^{5} (-1)^{k} \binom{5}{k}(10 - k)^9 \end{align*} where $\binom{5}{k}$ is the number of ways of excluding $k$ of the $5$ odd digits and $(10 - k)^9$ is the number of ways of filling the nine positions of the string with the remaining $10 - k$ decimal digits.