I come across the problem of all the choices of selecting a subset of $n$ objects.
When expressed in binary form, seems very clear that there are $2^n$ subsets.
But when using combination, it comes to this result:
$$ \sum_{x=0}^n \frac{n!}{(n-x)! x!} = \sum_{x=0}^n \binom{n}{x} . $$
How do we prove those two are equivalent? If they are indeed so?
$$ \sum_{x=0}^n \frac{n!}{(n-x)! x!} = 2^n $$
