Prove the number of symmetric relation is $2^{\frac{n^2+n}{2}}$ If the number of set $A$ is given. let $n(A)=n$.
Prove the number of symmetric relation from $A$ to $A$   is $2^{\frac{n^2+n}{2}}$
We know that number of  relations from $A$ to $A$ is $2^{n^2}$
this is obtained by, number of subsets of $A$x$A$ $=2^{n^2}$
every subsets of $A$x$A$  is the relation from $A$ to $A$
like this is any proof is there?

Given two distinct elements $x,y$ in $A$, you can either have both $(x,y)$ and $(y,x)$ in the relation or not.  You also can have $(x,x)$ or not.  The number of choices you make is the number of unordered pairs from $A$ including pairs of the same member.  There are $\frac {n^2+n}2$ of those.
