How do you compute the number of reflexive relation? Given a set with $n$ elements
I know that there is $2^{n^2}$ relations, because there are $n$ rows and $n$ columns and it is either $1$ or $0$ in each case, but I don't know how to compute the number of reflexive relation. I am very dumb. Can someone help me go through the thought process?
 A: A relation is reflexive if and only if every entry on the main diagonal of its matrix is $1$; that’s the only restriction. Fill in $1$’s on the diagonal, and you can put either $0$ or $1$ freely into every other entry in the matrix and have the matrix of a reflexive relation. Thus, the number of reflexive relations on a set of $n$ elements is $2^m$, where $m$ is the number of entries that are not on the diagonal. There are $n^2$ entries altogether, and $n$ of them are on the diagonal, so how many are not on the diagonal? And then how many reflexive relations are there?
A: Strange way you have to count the number of relations...A relation on a set $\,A\,$ is just a subset of the cartesian product $\,A\times A\,$, and if $\,|A|=n\Longrightarrow |A\times A|=n^2\Longrightarrow\,$ the number of subsets of $\,A\times A\,$ , i.e. $\,|P(A\times A)|\,$ , is $\,2^{|A\times A|}=2^{n^2}\,$...
Now, you need to count all the subsets of $\,A\times A\,$ that contain the diagonal $\,\Delta_A:=\{(a,a)\in A\times A\}\,$ , so...can you take it form here?
A: A relation $R$ on $A$ is a subset of $AXA$. 
If $A$ is reflexive, each of the $n$ ordered pairs $(a,a)$ belonging to $A$ must be in $R$. 
So the remaining $n^2-n$ ordered pairs of the type $(a,b)$ where $a!=b$ may or may not be in R. 
So each ordered pair has now 2 choices, to be in $R$ or to not be in $R$. 
Hence number of pairs = 2($n^2-n$) 
