, 2020

Solution:

We can obtain a good estimate by only counting terms of the form $\binom{a}{1}$, $\binom{a}{2}$, $\binom{a}{a-1}$, and $\binom{a}{a-2}$. The last two cases are symmetric to the first two, so we will only consider the first two and multiply by 2 at the end.

Since $\binom{a}{1}=a$, there are 90,000,000 values of $a$ for which $\binom{a}{1}$ has eight digits. Moreover, since $\binom{a}{2} \approx a^{2} / 2$, the values of $a$ for which $\binom{a}{2}$ has eight digits vary from about $\sqrt{2 \cdot 10^{7}}$ to $\sqrt{2 \cdot 10^{8}}$, leading to about $10^{4} \sqrt{2}\left(1-10^{-1 / 2}\right) \approx 14000 \cdot 0.69=9660$ values for $a$.

Therefore, these terms yield an estimate of 180019320, good enough for 13 points. Of course, one would expect this to be an underestimate, and even rounding up to 180020000 would give 16 points.