<s>[INST] How many numbers smaller than one milion? How many numbers smaller than 1 000 000 :
a) have digits in a non-decreasing order?
b) contains exactly three digits 9 and have an odd sum of numbers?
c) have digits in non-increasing order?
For the a. solution 
Is it like: 
$\binom{6+10-1}{10-1}$
Am I right?
Solution to the b.
So the first solution is exactly 999, 
for the 4-digits number $\binom{4}{1}$$\binom{4}{1}$ +1 ,
for the 5-digits number $\binom{5}{1}$$\binom{4}{1}$$\binom{4}{1}$$\binom{4}{1}$ 
 [/INST] Your answer for the first question is correct if the numbers are nonnegative integers.  As @skyking pointed out, we get the same result for the third question.  

How many nonnegative integers less than $1~000~000$ contain exactly three $9$s and have an odd digit sum?

We treat the nonnegative integer as a six-digit decimal string, by appending leading zeros to numbers with fewer than six digits as necessary.
There are $\binom{6}{3}$ ways to choose three of the six positions for the $9$s.  
Since  $3 \cdot 9 = 27$, the sum of the remaining three digits must be even.  To get an even sum, we must either use three even digits or one even digit and two odd digits when filling the remaining three positions.  
Three even digits:  We have five choices for each even digit, so this can be done in $5^3$ ways.
One even digit and two odd digits:  We have three ways to choose the position of the even digit and five ways to choose the even digit we place in that position.  Since we cannot use $9$, we have four ways to fill each of the remaining open positions with an odd digit.  Hence, there are $3 \cdot 5 \cdot 4^2$ such choices.  
Thus, the number of nonnegative integers less than $1~000~000$ that contain exactly three $9$s and have an odd digit sum is 
$$\binom{6}{3}\left[5^3 + 3 \cdot 5 \cdot 4^2\right]$$ 
Finally, since the number $0$ does not contain three $9$s, the same argument applies to positive integers.
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