Given , the cardinality of set = n
So consequently ,
No of entries in operation table(Cayley table) = n^{2}
And hence if we consider lower triangular or upper triangular half , we have : (n^{2} + n) / 2
And in an operation table , each entry can be filled in n ways by any one element out of given n elements of the set..
So no of ways we can fill the upper or lower triangular half = n^{(}^{n^2 + n)}^{/2}
So each of these is nothing but an instance of operation table of commutative operation as say (i,j) entry is filled in the table so (j,i) entry will also be the same hence the choice for (j,i) entry is constrained to 1 as we are concerned about commutativ operation table here..
Therefore,
No of possible binary operations which are commutative = n^{(}^{n^2 + n)}^{/2}