<s>[INST] Find number of positive integers less than $10^8$ with digit sum of $24$ 
Find number of positive integers less than $10^8$ with digit sum of $24$.

I thought I could use the argument $a+b+c+d+e+f+g+h = 24$ and then make the substitution $x_1 = 9-a, x_1 = 9-b,\ldots,x_8 = 9-h\ldots$ but then that would also count negative values of $a,b,\ldots,h$ so I am thinking there is a better way to count this. Or maybe my way could work and we subtract away the negative cases?
 [/INST] Generating function approach.
You want the coefficient of $x^{24}$ in:
$$\begin{align}(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9)^{8} &= \left(\frac{1-x^{10}}{1-x}\right)^{8}\\
&=(1-x^{10})^8\sum_{k=0}^\infty\binom{k+7}{7}x^k
\end{align}$$
We can ignore the higher terms of $(1-x^{10})^8 = 1-8x^{10}+28x^{20}-\cdots$ because they do no contribute to $x^{24}$.
So you get: $$\binom{24+7}{7}-8\binom{14+7}{7}+28\binom{4+7}{7}$$
This can also be seen as an inclusion-exclusion result. $\binom{24+7}{7}$ is the number of ways of writing $24$ as the sum of eight non-negative integers. For each position, we subtract the $\binom{14+7}{7}$ solutions where that position is $10$ or greater. But then we've subtracted some of the sums twice, so we have to add back the cases where two of the values are $10$ or greater.
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