Say I have a set $A = \{a,b,c \}$. I want to know how many equivalence relations are there. I know the answer is $5$ from wikipedia, but I am not sure if the technique I am using is right. I want to use a counting approach.
There is only one way to put all the elements in its own subsets, namely, $A \subseteq A$.
There is $\binom{3}{2}$ ways for two elements in one subset and one in another.
Then there is only one way for each element to be in its own subset.
All together we have $1+3+1 = 5$. Is this the correct way of thinking when we use counting to count the partitions for a set?
You have $9$ pairs in $A \times A$. You have to include $\Delta_A$ in $R$ to achieve reflexivity. So no choice there. Adding a pair $(x,y)$ to $R$ means you add $(y,x)$ too by symmetry.
Adding a single such "combo pairs" will give an equivalence relation (no real check for transitivity is needed yet; we get two $R$-classes), adding two such wil force the other ones in too. $A \times A$ also is an equivalence relation. We have $3$ relations with $2$ classes, $1$ with three classes ($\Delta_A$) and $1$ with a single class ($A \times A$).
So $5$ relations in total. For such a small set this is doable by hand; larger sets are harder, see Bell numbers.