Number of asymmetric relation Let $A$ be a set of $n$ elements. How can we calculate number of asymmetric relations on $A$? I googled for it and got the answer that the number is given by $3^{(n^2-n)/2}$, But I don't know how to derive it.
 A: Let us take an example.
$$A = \{1,2,3\}$$
$$A\times A = \{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}$$
We know that, $(1,1),(2,2),(3,3)$ won't come in the asymmetric relation.(3 elements)
Remaining elements of $A\times B$ can be formed a group.
$[(1,2),(2,1)]$, $[(1,3),(3,1)]$, $[(2,3),(3,2)]$ (3 groups)
From every group, we can choose elements in 3 different ways.


*

*choose 1st element.

*choose 2nd element.

*choose no elements.


That is, we can choose elements in $3^3$ different ways in this case.
Generalizing,
If there are $n$ elements in the set, the number of elements in the group is $\frac{\text{Total  number  of relations - reflexive  elements }}{2}=\frac{n^2 - n}{2}$
Each of these elements can be chosen in 3 ways. Therefore the number  of asymmetric  relations is $3^{\frac{n^2 - n}{2}}$ 
A: Given elements $a,b\in A$, either blank, blank, blank, or blank. If the relation $R$  on $A$ is asymmetric, only blank, …, or blank can occur. There are blank distinct unordered pairs of elements of $A$, so the total number of asymmetric relations on $A$ is blank.
