<s>[INST] 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?
 [/INST] 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.
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