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?
 A: An alternative method:
Choose the three locations for the $9$s: there are $\dbinom 63$ possibilities.
Then, among the remaining digits (not $9$s), choose any digit except $9$ for the leftmost place ($9$ possibilities), then choose any digit except $9$ for the middle place ($9$ possibilities).
Among the ways to fill these two digits, there are $4^2 + 5^2$ in which we choose two digits of the same parity and get an even sum, 
in which case we need to choose an even digit for the last open place ($5$ choices).
There are $2\times4\times 5$ ways in which to get an odd sum, in which case we need one more odd digit ($4$ choices, since $9$ is excluded).
So altogether we have
$$
\binom 63 \left(5(4^2 + 5^2) + 4(2\times4\times 5) \right) = 7300.
$$
A: 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.
A: There are $\frac{6!}{3!3!}$ to choose where the 9's are and where the even numbers are. However, even if we know that the three even numbers are equal, there are $5$ ways to choose if it is $0,2,4,6$ or $8$. (Note that we don't care about leading zeros, because they don't affect the sum of digits.) So there are $5\frac{6!}{3!3!}$ such ways. When we have two different even numbers, surely there is $\frac{6!}{3!2!}$ ways to separate 9's positions, and two even number positions, but there are also $5*4=20$ ways to choose actual even numbers. In the case of different even numbers we have $5*4*3=60$ ways to choose even numbers for the first, second and third non-9 digit and $\frac{6!}{3!3!}$ ways to choose 9's places before. So in the case of even digits there are
$$5\frac{6!}{3!3!}+20\frac{6!}{3!2!}+60\frac{6!}{3!3!}=\frac{6!}{3!}(5/6+10+10)=20*(5+60+60)=2500$$
However, note that in fact separating three cases doesn't give any advantage to us. We would better say that there is $\frac{6!}{3!3!}$ ways to choose the places of $9$'s and then $5^3$ ways to choose even numbers for other three places: just five choices for each place, independently from the others. And we obtain $$125\frac{6!}{3!3!}=4*5^4=2500$$
The same, all is ok.
For the case of two odds we can do the following: choose 9s places in $\frac{6!}{3!3!}$ ways, then choose in 3 ways where is even and in 5 ways what is it. For two odds we can choose out of $1,3,5,7$, so we have $4^2$ choices. All in all, for the odd case,
$$\frac{6!}{3!3!}*3*5*16=20*15*16=4800$$
And finally, $2500+4800=7300$ is the answer.
A: This is a slight variation on David K's answer. It's the same except for how we deal with the last digit. I'd hoped to avoid cases by exploiting symmetry, but it turned out I couldn't completely avoid them.


*

*Choose the positions of the three $9$'s (there are $6 \choose 3$ ways to do this)

*Moving left to right, choose two of the remaining three digits (there are $9^2$ ways to do this)

*Choosing the last digit, call it $k$, it's a little tough to make sure the digit sum is odd. But we can exploit symmetry a bit: 


*

*If you choose any one of $1, 2, 3, 4, 5, 6, 7, 8$, then if you'd chosen $9 - k$ instead you'd wind up with a number whose digit sum is of the opposite parity (e.g., if you choose $6$ and the sum is even, then having chosen $9 - 6 = 3$ would leave you with an odd digit sum). There are $8$ possibilities here, half of them will be acceptable whether the sum without this digit is even or odd.

*It remains how to deal with $0$ (since we've chosen our three $9$'s). Unfortunately, I can't think of a better way than messing with cases of the two non-$9$ chosen digits. We can choose $0$ as the final digit if and only if the two numbers have an even sum. So the two can both be even ($5^2$ ways), or both odd ($4^2$ ways).
So our total is ${6 \choose 3} \cdot \big[ (9^2 \cdot 4) +(4^2 + 5^2)\big]$ (not surprisingly also $7300$), where the sum corresponds to whether our "final" digit is nonzero (in which case we can choose the "first" two freely) or not. 
A: Some brute force which quickly produces, unsurprisingly, 7300. (Python)
N= 1000000

count=0
for i in range(1,N):
    nines = 0
    dsum = 0
    for c in str(i):
        if c == '9': nines += 1
        dsum += int(c)
    if dsum % 2 == 1 and nines == 3: count += 1
print count

A: Assuming leading zeros allow us to assume all numbers have six digits you need.
A) 1 non-nine odd, two different evens, 3 nines
plus 
B) 1 (non-nine odd), two same evens, 3 nines
plus
C) 3 non-nine odds, all different, 3 nines
plus
D) 3 non-nine odds, two the same, 3 nines
E) 3 non-nine odds, all the same, 3 nines.
There are $4$ odds to choose from and $5$ evens.
A) is $4*5*4*{3\choose 1}{6 \choose 3}$.  That is, $4$ choices for the non-nine odd, $5$ for the first even, $4$ for the second even.  There are ${3\choose 1}$ ways to arrange the odd digit among the two even digits.  And there are  ${6\choose 3}$ ways to place the three nines among six digits.
B) is $4*5{3\choose 1}{6 \choose 3}$.  That is, $4$ choices for the first non-nine odd, $5$ choices for the two evens. There are ${3\choose 1}$ ways to arrange the odd digit among the two even digits.  And there are  ${6\choose 3}$ ways to place the three nines among six digits.
C) is $4*3*2{6\choose 3}$
D) is $4*3{3 \choose 2}{6\choose 3}/2$
E) is $4*{6 \choose 3}$
There is an alternative.
There is ${6 \choose 3}$ ways to arrange three $9$s is six positions.  There are $9^3$ possible options for the remaining three digits.  $5^3$ of them are all evens (as there are $5$ choices of non-nine evens).  $4^3$ are all odd (as there are $4$ choices of non-nine odds).  $4*5^2{3\choose 1}$ of those have $1$ odd and $2$ evens.  $4^2*5{3\choose 1}$ of those have $2$ odds and $1$ even.
Detour: Verify the $9^3 = 5^3 + 5^2*4*{3\choose 1} + 5*4^2*{3\choose 1} + 4^3$
$9^3 = (5+4)^3 = 5^3 + 5^2*4*{3\choose 1} + 5*4^2*{3\choose 1} + 4^3$ is just the binomial theorem.  Nothing to verify.
End of detour.
Of those $5^3$ (three evens) and $5*4^2*3$ (two odds, one even) have an even sum of digits.  So when three $9$s are added the sum will be odd.
So there are ${6 \choose 3}(5^3 + 5*4^2*3)$ six digit numbers with exactly three $9$s and an even sum of digits.
