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?

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.
