How to proof that more than half binary algebraic operations on a finite set are non-commutative?

We know that if $S$ is a set and $|S|=n$, then there are $n^{n^{2}}$ binary algebraic operations, right?

The cardinality of $|S|=n$ and cardinality of $|S\times S|=n^{2}$. Also, the number of all mirrors (or all functions) between these two sets will be $n^{n^{2}}$. We know that these functions are by other words: algebraic binary operations.

But how do I to prove that there are $n^{\frac{n^{2}+n}{2}}$ commutative binary operations on $S$?
How do I prove that more than half binary algebraic operations on a finite set are non-commutative operations?

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Let $E=\{e_1,e_2,…e_n\}$, then binary operations,$\Delta$, defined on $E$ are functions from cartesian product $E$X$E$ to E. The set $E$X$E$ has cardinality $n^2$. There are $n$ ordered pairs in $E$X$E$ where both elements are the same for each distinct ordered pairs. From $E$ We can select $\begin{pmatrix}
n \\
2 \\
\end{pmatrix}$ ordered pairs where the pairs are distinct; if we reverse the order of each of these pairs we can have $\begin{pmatrix}
n \\
2 \\
\end{pmatrix}$ more ordered pairs. The total number of ordered pairs in $E$X$E$ is $n^2= \begin{pmatrix}
n \\
2 \\
\end{pmatrix}+ \begin{pmatrix}
n \\
2 \\
\end{pmatrix}+ n$.

For our purpose (commutative binary operations) we are only interested in $ \begin{pmatrix}
n \\
2 \\
\end{pmatrix}+ n$ ordered pairs since the reverse of each ordered pair must point to the same element in E as that ordered pair. Therefore, we can think of our problem as: how many binary operations from a set of cardinality $ \begin{pmatrix}
n \\
2 \\
\end{pmatrix}+ n$ to E of cardinality $n$. Therefore, we have a total of $n^{\frac{n^{2}+n}{2}}$ commutative binary operations.

QED

I want to rewrite this with more rigor and mathematical notations.

Note that a commutative binary operation on the set $S = \{ 1 , 2 , \ldots , n \}$ is completely determined by how it acts on the following set of pairs: $$\{ \langle i,j \rangle : 1 \leq i \leq j \leq n \}.$$ How many elements are there of this set? The rest should be a simple counting argument.

HINT: Let $S=\{s_1,s_2,\ldots,s_n\}$. If $\otimes$ is a commutative binary operation on $S$, then $\otimes$ is completely determined by the values of $s_j\otimes s_k$ for $1\le j\le k\le n$, and those values can be chosen independently of one another.