How many relations are there on a set with n elements that are symmetric and a…

Symmetric relations allow free choice for n diagonal pairs (2^n) and n(n-1)/2 off-diagonal pairs (2^(n(n-1)/2)). Total is 2^n * 2^(n(n-1)/2) = 2^(n(n+1)/2). Reflexive relations fix all n diagonal pairs. Off-diagonal choices remain 2^(n(n-1)/2). Total is 2^(n(n-1)/2).