We first count the number of ways to produce an even number. The last digit can be any of $2$, $4$, or $6$. So the last digit can be chosen in $3$ ways.
For each such choice, the first digit can be chosen in $6$ ways. So there are $(3)(6)$ ways to choose the last digit, and then the first.
For each of these $(3)(6)$ ways, there are $5$ ways to choose the second digit. So there are $(3)(6)(5)$ ways to choose the last, then the first, then the second.
Finally, for each of these $(3)(6)(5)$ ways, there are $4$ ways to choose the third digit, for a total of $(3)(6)(5)(4)$.
Similar reasoning shows that there are $(4)(6)(5)(4)$ odd numbers. Or else we can subtract the number of evens from $840$ to get the number of odds.
Another way: (that I like less). There are $3$ ways to choose the last digit. Once we have chosen this, there are $6$ digits left. We must choose a $3$-digit number, with all digits distinct and chosen from these $6$, to put in front of the chosen last digit. This can be done in $P(6,3)$ ways, for a total of $(3)P(6,3)$.
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
You are right. Here's another way to see it. The 8 digit numbers are those from $10^7$ to $10^8-1$ inclusive, so the number of them is $10^8-10^7=9\times 10^7$.