How many numbers smaller than one million, their sum of digits is at least 20? 
How many numbers smaller than one million, their sum of digits is at least 20?

My attempt:
Since I don't know how to handle the "at least" part, I'll be using a complement:
The general case is all the digit combination up to one million: $9\cdot10^5$
Complement: the sum is at most 20.
Now we need to find all the solutions for $x_1+...+x_6=n : n\in[0,20]:x_i\in[0,9]$
So I found the generating sums function: $\frac{(1-x^{10})^6}{(1-x)^7}$
And the coefficients I got from that for $x^{20}$ are: $\binom{20+6}{6}-6\binom{10+6}6+\binom 6 2$
So the final answer is: $9\cdot10^5-\left(\binom{20+6}{6}-6\binom{10+6}6+\binom 6 2\right)=717803$
But I'm not sure I got the general case right, is it really all the 6 digit combinations?
 A: To simplify the calculations, we will work with nonnegative integers.
There are $1,000,000$ nonnegative integers less than $1,000,000$ since $0$ is nonnegative.
If the sum of the digits of a nonnegative integer is not at least $20$, then the sum of its digits is at most $19$.
A number less than $1,000,000$ has at most six digits.  We can append leading zeros to a number with fewer than six digits to represent it as a six-digit number.  For instance, we regard $473$ as $000473$.  
Thus, we wish to exclude nonnegative integers whose digits satisfy the inequality
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 \leq 19$$ 
subject to the restrictions that $x_k \leq 9$ for $1 \leq k \leq 6$.
To handle the inequality, let $d = 19 - (x_1 + x_2 + x_3 + x_4 + x_5 + x_6)$. 
Then $d$ is a nonnegative integer and 
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + d = 19$$
If there were no restrictions on the size of $x_k$ for $1 \leq k \leq 6$, then the number of solutions of this equation in the nonnegative integers would be equal to the number of ways we can place six addition signs in a row of $19$ ones, which is
$$\binom{19 + 6}{6} = \binom{25}{6}$$
However, we have counted solutions in which one of the $x_k$'s is greater than $9$.  Note that at most one $x_k$ can exceed $9$ since $10 + 10 = 20 > 19$. 
Suppose that $x_1 \geq 10$.  Let $y_1 = x_1 - 10$.  Then $y_1$ is a nonnegative integer and 
\begin{align*}
x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + d & = 19\\
y_1 + 10 + x_2 + x_3 + x_4 + x_5 + x_6 + d & = 19\\
y_1 + x_2 + x_3 + x_4 + x_5 + x_6 + d & = 9 
\end{align*}
This equation has 
$$\binom{9 + 6}{6} = \binom{15}{6}$$
solutions in the nonnegative integers.  Since any of the six $x_k$'s could exceed $9$, the number of nonnegative integers less than $1,000,000$ such that the sum of the digits is less than $20$ is 
$$\binom{25}{6} - \binom{6}{1}\binom{15}{6}$$
a total that includes the number $0$.  
Thus, the number of nonnegative integers less than $1,000,000$ such that the sum of the digits is at least $20$ is 
$$1,000,000 - \left[\binom{25}{6} - \binom{6}{1}\binom{15}{6}\right]$$
