How many of the 6 digit strings 000000 through 999999 have digits which sum to 27? with Build up counting. I know we can solve this with 
$${27 + 5 \choose 5} = {32 \choose 5}$$ 
but how would I do this $\textbf{strictly with the build up counting method?}$ 
I was thinking to have the base case $Q(3) = 1$ since this would be '$999$' and less than 3 digits is not possible. 
And then we would develop a recursion for strings of length greater than $3$. But I cannot seem to figure this out. 
 A: Here's a solution that uses generating functions:
$$\text{Let }p(x)=\sum_{i=0}^9x^i=1+x+x^2+\ldots+x^8+x^9.$$
The answer you are looking for will be the coefficient of $x^{27}$ in $\left(p(x)\right)^6$.
I used a TI-92 calculator to find the coefficients of $\left(p(x)\right)^6$. It took nearly a minute, but it eventually got an answer of $55,252$.
Since you asked for a recursive formula, I'll say a little bit about how you could use the above idea to find such a formula.
Define $a(n,k)$ to be the number of $k$-digit strings whose digits add up to $n$. Define $p(x)=1+x+x^2+\ldots+x^8+x^9$, as above, and note that $a(n,k)$ is also the coefficient of $x^n$ in $\left(p(x)\right)^k$.
Now for $k\ge2$, $$\left(p(x)\right)^k=p(x)\cdot\left(p(x)\right)^{k-1}.$$
Using the fact that $a(n,k)$ is the coefficient of $x^n$ in this product, we get the following recurrence relation:
$$a(n,k)=\sum_{i=0}^9 a(n-i,k-1).$$
So to find the number of $6$-digit sequences that whose digits add up to $27$, we could start by noting the following:
$$a(27,6)=\sum_{i=0}^9 a(27-i,5).$$
Then each of $a(27,5)$, $a(26,5)$, ..., $a(18,5)$ could be expressed in terms of $a(27,4)$, $a(26,4)$, ..., $a(9,4)$. Etc.
