Number of antisymmetric relationships in set Let's say I have $20$ elements in set A. 
I want to know the number of antisymmetric relations that are possible in $A \times A.$
antisymmetric $\iff a R b \wedge b R a \implies a = b$
The way I think about this is:


*

*There's $2^{20}$ number of possible relations that have or don't have a reflexive relationship between each member.

*There's about $2^{n(n-1)/2}$ number of possible connections between nodes (without counting the reflexive connections). Of those connections there are three possible ways (vs 1) that are anti-symmetric. (I'm thinking about this problem as a graph or geometry problem).
So the number of anti-symmetric relationships is something like:
$2^{n} 3^{n(n-1)/2}$ 
(with $n = 20$)
Is this correct?
And also, is this the default line of thought regarding this problems? Maybe there's a more straightforward way to think about it.

Thinking of it as a graph is a good idea.  You have $20$ vertices.  For each pair, you can have one of three choices, no edge meaning neither direction is related or one of two directions of directed edge meaning one is related to the other.  There are $\frac 12 20(20-1)=190$ pairs, so there are $3^{190}$ antisymmetric relations.  Then as you say you can choose the self-related elements in $2^{20}$ ways, so the total is $2^{20}3^{190}$
