I can calculate this by

$$20+\left(\begin{array}{c}{20} \\ {2}\end{array}\right)+\left(\begin{array}{c}{20} \\ {3}\end{array}\right)+\ldots+\left(\begin{array}{l}{20} \\ {20}\end{array}\right)$$

But this does not correspond to any of the answers (it's a multiple choice). 

Q1: Are there a compact way to calculate this via some theorem?

Q2: This is a combination and not a permutation, am I right?