For a finite set $S$ there are $\vert S\vert!$ orderings of its elements.
What is the cardinality of all orderings of $\mathbb{N}$? What would $$\vert \mathbb{N}\vert!$$ mean? Is it $\prod_{n\in\mathbb{N}} \aleph_0=\aleph_0^{\aleph_0}=\aleph_1$?
Wikipedia says the "equivalence classes of well-orderings of the natural numbers" has cardinality $\aleph_1$. But I can't put together in my mind the connection between well-orderings of $\mathbb{N}$ and all orderings of $\mathbb{N}$. (Are there non well-ordered orderings of $\mathbb{N}$?)
This leads to wondering what the cardinality of all orderings of $\mathbb{R}$ is. (Actually my thought process went in the other direction, from $\mathbb{R}$ to $\mathbb{N}$.) I think knowing the answer to the above question will teach how to determine what $$\vert \mathbb{R}\vert !$$ means, if anything.
I will be happy if you assume the continuum hypothesis to be true for the purposes of this question, but any level of completeness is welcome.
First thing first, you don't need the continuum hypothesis if you are willing to use the provable cardinality of $\Bbb R$ being $2^{\aleph_0}$, also denoted by $\frak c$ often.
Now. One can easily prove that $\Bbb N$ has $\aleph_0^{\aleph_0}=2^{\aleph_0}=\frak c$ permutations. Given any permutation $f\colon\Bbb N\to\Bbb N$, we can define an order $<_f$ on $\Bbb N$ by taking $f(n)<_f f(m)\iff n<m$. Since any two distinct permutations induce two different orders, this means that there are at least $2^{\aleph_0}$ ways to linearly order $\Bbb N$ which. And there can't be any more than that too, because there are only $2^{\aleph_0}$ relations on $\Bbb N$ to begin with.
Similarly with $\Bbb R$. We can quite easily prove that there are $2^{2^{\aleph_0}}$ permutations of $\Bbb R$, each inducing a different ordering, so there are $2^{2^{\aleph_0}}$ ways to linearly order the real numbers.
As a side remark, $\aleph_1$ is defined to be the cardinality of all countable ordinals, which is exactly the different isomorphism types of countable well-orders. But even then, there are $2^{\aleph_0}$ ways to order a countable set in a given order type.