Let $G$ be a group and $a,b,c,\ldots$ elements of $G$. How many different ways are there to multiply elements in $G$, preserving the order?
I mean, two elements ----> 1 way: $ab$
three elements ----> 2 ways: $(ab)c$; $a(bc)$
four elements ----> 5 ways: $((ab)c)d$ ; $(a(bc))d$; $(ab)(cd)$; $a(b(cd))$; $a((bc)d)$
$n$ elements----> how many ways?
Note: we assume we can multiply only pairs of elements, so expressions like $abc$ don't make sense.
Basically, you're asking how many different binary trees with $n$ unlabeled nodes are there, as each multiplication expression yields such a tree and vice versa.
Wikipedia has the solution: $C_n = \frac{1}{n+1}{2n\choose n}$, (n+1) is the number of elements, $C_n$ is the $n$th Catalan number.
However, this number does not agree with the image size of the map which takes $n$ elements from an abelian group $G$ and multiplies them, as there are inverse elements: $ab$ may equal $cd$ for $\{a, b, c, d\} \subseteq G, |\{a, b, c, d\}| = 4$ if $a = b^{-1}$ and $c = d^{-1}$.