N distinct elements, is a permutation or all possible subsets larger?

Question

Given N distinct elements (where N is large), how do I determine if a permutation of it is larger, or if all possible subsets is larger?

I eyeballed it and it seems like the permutations would be bigger. It seems like this subset is largest: N choose N/2, but this subset is kind of close to $2^{N/2}$, which is clearly a lost smaller than $N!$. Not sure how to prove this though

Answer 1

If you mean whether the size of the set of all possible permutations of an N-element set is bigger than the size of the set of all possible subsets, i.e. whether $N!>2^N$ then the answer is yes, if $N>3$. That it is wrong for $N=1,2,3$ is a simple computation. The actual statement follows by induction:

1. The induction start: we take $N=4$ and obtain $N!=24>16=2^4$
2. Now suppose the statement holds for $N-1$, then since clearly $N>2$ we have $$N!=N(N-1)!>2(N-1)!>2\cdot 2^{N-1}=2^N$$ where the last inequality follows from the induction hypothesis.

Answer 2

The orders are $2^n$ and $n!$. But $n!\gt 2^n$ for $n\gt3$, by inspection. So there are more permutations, in this case.