Given that $|A|=n$, how many asymmetric relations are there? ($A\to A$)

Apparently answer is $3^{\frac{n^2-n}{2}}2^n$ (???)

I understand where $3^{\frac{n^2-n}{2}}$ came from, but I don't understand about $2^n$ part... Can someone explain?

Question

Given that $|A|=n$, how many asymmetric relations are there? ($A\to A$)

Apparently answer is $3^{\frac{n^2-n}{2}}2^n$ (???)

I understand where $3^{\frac{n^2-n}{2}}$ came from, but I don't understand about $2^n$ part... Can someone explain?

perl · Answer

can you quickly define an asymmetric relation (just to review )

perl · Answer

symmetric relation is
if a R b  -> bRa

geerky42 · Answer

Definition of Asymmetric given: For all $a,b\in A$, whenever $a$ is related to $b$, then $b$ is not related to $a$.

michele_laino · Answer

I think that 2^n is, because every relation have to be counte two times since it is asymmetric, namely from domain to range and from range to domain

michele_laino · Answer

I think the same relation, since is asymmetric must be counted two times. I tried to verify your formula by mathematical induction principle

geerky42 · Answer

Maybe stupid question, but in that case, why not just $\large 2\cdot3^{\frac{n^2-n}{2}}$?

Why is $2$ raised to the power of $n$ ?

geerky42 · Answer

Yeah, stupid question.

michele_laino · Answer

becaus for every point that I add to A, I can define  2^cardinality of A relations