The problem is easy:
How many two digits number exist which are power of two?
While it is not difficult to figure out just using $2^n$ I am more confused what combinatory formula can fit this problem (my professor requires formula and proof), Can someone help here?
We want the largest integer $n$ satisfying $2^n \le 99$. If we drop the integer requirement and take logarithms, we have $$n \ln(2) \le \ln(99)$$ so $$n \le \frac{\ln(99)}{\ln(2)} \approx 6.63$$ Now impose the integer requirement and we have $n=6$.
(Personally, I don't see anything wrong with simply observing that $2^6 < 99$ and $2^7 > 99$, but maybe your teacher wouldn't like that.)