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:


 * $$N = n^{\frac {n \left({n+1}\right)}2}$$

Proof
Let $$\left({S, \circ}\right)$$ be a groupoid.

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

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

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

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:
 * $$\binom n 2 = \frac {n \left({n - 1}\right)} 2$$

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 $$\frac {n \left({n - 1}\right)} 2 + n = \frac {n \left({n+1}\right)} 2$$.

The result follows from Cardinality of Set of All Mappings.

Comment
The number grows rapidly with $$n$$:

$$\begin{array} {c|cr} n & \frac {n \left({n+1}\right)}2 & n^{\frac {n \left({n+1}\right)}2}\\ \hline 1 & 1 & 1 \\ 2 & 3 & 8 \\ 3 & 6 & 729 \\ 4 & 10 & 1 \, 048 \, 576 \\ \end{array}$$

and so on.