TITLE: How to find the number of anti-symmetric relations?
QUESTION [12 upvotes]: I know that given a set $A = \{1, 2, 3, ... , n\}$, the total number of relations on $A$ is $$2^{n^2}$$ The number of reflexive relations is $$2^{n^2 - n}$$ The number of symmetric relations is $$2^{{n+1}\choose 2}$$ But how can I find the number of anti-symmetric relations? With a small set, say $n = 4$, it can be easy to just brute force it. Is there another way (perhaps using the inclusion-exclusion principle?)

REPLY [5 votes]: Maybe a different way to count the number of antisymmetric relations (I nowhere found this approach, so I post it here). Every relation on $n$ elements could be viewed as a boolean matrix $A = \{(a_{ij})\}_{i,j=1,\ldots, n}$ of size $n \times n$. So we have to count all those boolean matrices that correspond to antisymmetric relations. 
First on the diagonal we are free to choose, hence we have $2^n$ possibilities. Now on the non-diagonal entries, consider first the upper right triangular part, i.e. all entries $a_{ij}$ with $i < j$. If for some entry $a_{ij} = 1$, then $a_{ji}$ must be $0$. Otherwise it must be zero itself, but in that case, the other value $a_{ji}$ could be choosen freely.
So we count by first deciding how many $1$'s we have in the upper right triangular part above the diagonal, and then we can freely choose the remaining values in the lower left corner beneath the diagonal which did not correspond to a one, i.e. $a_{ji} \in \{0,1\}$ for $a_{ij} = 0$. Hence we have
$$
 \sum_{k=0}^{\binom{n}{2}} \binom{\binom{n}{2}}{k} 2^{\binom{n}{2} - k}
$$
where the sum is over every matrix after choosing how many ones it will have in the upper right triangular part, which has $\binom{n}{2}$ values in total, $\binom{\binom{n}{2}}{k}$ count the ways we can distribute exactly $k$ ones among these $\binom{n}{k}$ entries, and $2^{\binom{n}{2} - k}$ count the ways we can assign values to the remaining $\binom{n}{2} - k$ entries in the lower left triangular part. So in total we have
$$
 2^n \cdot \sum_{k=0}^{\binom{n}{2}} \binom{\binom{n}{2}}{k} 2^{\binom{n}{2} - k}
$$
antisymmetric relations, and as
$$
 3^{\binom{n}{2}} = (1 + 2)^{\binom{n}{2}} = \sum_{k=0}^{\binom{n}{2}} \binom{\binom{n}{2}}{k} 2^{\binom{n}{2} - k}
$$
our solution is consistent with the previous ones.