In Riemann rearrangement theorem, are the number of possible arrangements that converge to a given sum countably or uncountably infinite?
If there are uncountably infinite does that imply that cardinality of number of rearrangements is greater than cardinality of $\mathbb R$ i.e. $\aleph_1$
There are uncountably many different rearrangements that converge to any given sum.
If you have one rearrangement that gives you a given sum, you can create uncountably many more by selecting an arbitrary subset $A\subseteq \mathbb N$ and then swapping term number $2n$ and $2n+1$ iff $n\in A$. This doesn't change the sum, and since there are uncountably many possible $A$ to choose from, you get uncountably many rearrangements.
(Some slightly fancier footwork will be required if there are terms that appear more than once, and you don't consider interchanging them to produce a different rearrangement, but this can be overcome too).
If there are uncountably infinite does that imply that cardinality of number of rearrangements is greater than cardinality of $\mathbb R$ i.e. $\aleph_1$
The number of possible rearrangements is always the same as the cardinality of $\mathbb R$, namely $2^{\aleph_0}$. (Note that this cardinality is not necessarily $\aleph_1$ -- the claim that $|\mathbb R|=\aleph_1$ is the continuum hypothesis, which cannot be proved or disproved from the usual axioms of set theory).
This is not in conflict with the fact that there are $|\mathbb R|$ different rearrangements that give each of the $|\mathbb R|$ possible sums, because $|\mathbb R|\times|\mathbb R|=|\mathbb R|$.