Imagine that you choose your $10$ options sequentially. There are $20$ possibilities for the first option. After you’ve chosen it, $19$ possibilities remain for your second option. Continuing to reason in this fashion, we see that there are
$20\cdot19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11$
ways to choose $10$ options in sequence; this number can be more compactly written $\dfrac{20!}{10!}$.
However, we’ve counted each possible set of $10$ options many, many times: we’ve counted it once for every possible order in which we could have picked it. How many times is that?
Suppose that we have a set of $10$ things, and we line them up. We can choose any of the $10$ to go first; once that’s been done, we can choose any of the $9$ remaining objects to go second; and so on. Thus, we can arrange the $10$ objects in any of $10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=10!$ orders.
Going back to the original problem, this means that the figure of $\frac{20!}{10!}$ counts each $10$-option set $10!$ times, once for each of orders in which we could have selected it, i.e., once for each of its possible permutations. The number of different sets of $10$ options is therefore only $\frac1{10!}$-th of $\frac{20!}{10!}$, or $\frac{20!}{10!10!}$. This number is the binomial coefficient
$\binom{20}{10}=\frac{20!}{10!10!}\;.$
The same reasoning applied to the general problem of counting the $k$-element subsets of a collection of $n$ things leads to the binomial coefficient
$\binom{n}k=\frac{n!}{k!(n-k)!}\;.$