Factorial of number of permutations of an uncountably infinite set

Question

Is the factorial of an uncountably infinite set also uncountably infinite? And, if so, is it a larger infinity?

For context, I'm working on a problem in which I need to find the number of permutations of a power set, ℘{F}. {F} is either finite or countably infinite. I understand the number of permutations, in this case, is equal to (2^|F|)!. So, I was wondering if (2^|F|)! is well-defined when |F| is countably infinite.

Thanks.

Answer 1

If $A$ is any infinite set, there are exactly as many permutations of $A$ as there are subsets of $A$. This is easy to see if you know

1. That $|A|=|A\times A|$. This requires the axiom of choice in general, but if $A$ is equinumeraous to $\mathbb N$ or $\mathbb R$ there are elementary proofs that don't depend on AC.
2. The Cantor-Bernstein theorem.

First, there are at least as many permutations of $A\times A$ as there are subsets of $A$. Namely, choose one nontrivial permutation $\sigma$ of $A$, and then for each subset $B\subseteq A$ we can consider the permutation $$ f_B(x,y) = \begin{cases} (x,\sigma(y)) & \text{if }x \in B \\ (x,y) & \text{otherwise} \end{cases}$$ It ought to be evident that different $B$ give different $f_B$.

Second, there are at least as many subsets of $A\times A$ as there are permutations of $A$. Namely, each permutation of $A$ is a subset of $A\times A$ (but not all such subsets are permutations).

Answer 2

There are $2^{\mathfrak c}$ permutations. The number of permutations is less than or equal to the number of continuum length sequences of reals, which is $\mathfrak c^{\mathfrak c}$. Given one permutation, which AC guarantees you exists, as a well order, you can find $2^{\mathfrak c}$ by splitting it into pairs, taking all the binary strings of length $\mathfrak c$ and swapping pairs that correspond to $1$s.