Count of Commutative Binary Operations on Set

Theorem
Let $S$ be a set whose cardinality is $n$.

The number $N$ of possible different commutative binary operations that can be applied to $S$ is given by:


 * $N = n^{\frac {n \paren {n + 1} } 2}$

Proof
Let $\struct {S, \circ}$ be a magma.

From Cardinality of Cartesian Product, there are $n^2$ elements in $S \times S$.

The binary operations $\circ$ is commutative :
 * $\forall x, y \in S: x \circ y = y \circ x$

Thus for every pair of elements $\tuple {x, y} \in S \times S$, it is required that $\tuple {y, x} \in S \times S$.

So the question boils down to establishing how many different unordered pairs there are in $S$.

That is, how many doubleton subsets there are in $S$.

From Cardinality of Set of Subsets, this is given by:
 * $\dbinom n 2 = \dfrac {n \paren {n - 1} } 2$

To that set of doubleton subsets, we also need to add those ordered pairs where $x = y$.

There are clearly $n$ of these.

So the total number of pairs in question is:
 * $\dfrac {n \paren {n - 1} } 2 + n = \dfrac {n \paren {n + 1} } 2$

The result follows from Cardinality of Set of All Mappings.