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}$ 
 A: For the question (b) one way to consider it is that you have to select $3$ slots where you would put the $9$'s and the rest of the digits would need to come from a three-digit even number without $9$'s.
The number of the later can be found by counting the number of even and odd $k$-digit numbers without $9$s recursively. You have immediately that $o_1=4$ and $e_1=5$ (found by direct counting of $1,3,5,7$ and $0,2,4,6,8$ respectively). The recursion formula is then $o_{k+1} = 4e_k + 5o_k$ and $e_{k+1} = 5e_k + 4o_k$ (in similar way). We can use this to see that $e_3=365$ by stepping the formula. We could also derive a closed form expression $e_k = (9^k+1)/2$ and $o_k = (9^k-1)/2$.
This gives us the number (being $365$ times the number of ways we can select $3$ slots out of $6$):
$$365 \binom{6}{3}$$

The (a) and (c) questions are basically the same. The number is the number of ways we can select multiplicities of digits summing up to $6$ (once you know the number of $0$s, $1$s and so on there's only one way to form a non-decreasing sequence of them). This is equivalent to put $6$ tally-sticks and $9$ separators in a sequence (the number of tally-sticks between the separators is the number of corresponding digit). This basically means the ways to select $6$ slots (for the tally-sticks) out of $15$ or as you found:
$$\binom{15}{6} = \binom{15}{9}$$
A: 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.
