How many numbers smaller than $10^6$ contain exactly three '$9$'s and have an odd sum of digits? i came up with an idea that i choose 
Even Even Even 9 9 9 - and i sort it in ways such that 
all even are the same - $\frac{6!}{3!*3!}$
two are the same - $\frac{6!}{3!*2!}$
all are different - $\frac{6!}{3!}$
other one is that 
Odd Odd Even 9 9 9 
odd are the same - $\frac {6!}{3!*2!}$
all three are differnt - $\frac{6!} {3!}$
and my result is a sum of all this options.
Am i right?

With your idea:


*

*Three even digits and the 9:


*

*The same even digit every time (5 choices: $0, 2, 4, 6, 8$ and permutation): $5 \cdot \frac{6!}{3! \cdot 3!}$

*Two even digits are the same: $5 \cdot 4 \cdot \frac{6!}{3! \cdot 2!}$

*The three even digits are different: $5 \cdot 4 \cdot 3 \cdot \frac{6!}{3!}$


*Two odd digits, one even (4 choices for the odd digit: $1, 3, 5, 7$):


*

*The same odd digit : $\underbrace{5}_{even} \cdot \underbrace{4}_{odd} \cdot \frac{6!}{3! \cdot 2!}$

*Odd digits are different:  $\underbrace{5}_{even} \cdot \underbrace{4 \cdot 3}_{odd} \cdot \frac{6!}{3!}$



The answer is the sum of all cases.
