Number of transitive relations We have a set $A$ with cardinality $n$. How to find the number of transitive relations on $A$? 
Also how do we get the following results?
Number of reflexive relations on $A=2^{n^2-n}$
Number of symmetric relations on $A=2^{\frac{n(n+1)}{2}}$

The problem of finding the number of transitive relations on a set of n elements is non-trivial. The number of relations defined on the set itself grows exponentially ($2^{n^2}$)
For finding the other two, lets consider a matrix form of representing relations (assume rows & columns are ordered by the elements - where a 1 corresponds to existence of an element (ordered pair) & a 0 corresponds to non-existence).
On a set n elements, the matrix is a square matrix with n rows & n columns (where each element corresponds to an ordered pair of elements from the set).
Make the following observations -

All of the diagonal elements correspond to Reflexive Relation.
The mirror elements across the diagonal correspond to Symmetric Relation.
Total no of diagonal elements is $n$
Total no of non-diagonal elements is $n^2 - n$
Total no of mirror-element pairs is $\frac{n^2 - n}{2}$

Now, for a Relation to be Reflexive, all of the diagonal elements must be 1, the other elements may or may not exist (either 0 or 1).
In Symmetric Relations, all of the mirror elements occur in pairs i.e., either both 1 or both 0, the diagonal elements may or may not exist (either o or 1).
Counting this way, its clear that the number of Reflexive relations is $2^{\frac{n^2 - 2}{2}}$ and the number of Symmetric Relations is $(2^n)(2^{\frac{n^2 - n}{2}})$ which is equal to $2^{\frac{n^2 + n}{2}}$
