Solution
= Solution
In a <symmetric relation>, each of the $n$ diagonal pairs $(a,a)$ can be chosen independently. For each of the $\binom n2$ unordered pairs of distinct elements, either both ordered pairs $(a,b),(b,a)$ occur or neither occurs. Every such choice is free and produces a distinct <symmetric relation>. Hence
$$
\boxed{2^n2^{\binom n2}=2^{n(n+1)/2}.}
$$
For the <empty set> this correctly gives one <binary relation>, the empty one.