A={1,2,3}
We know that for a reflexive relation (a,a)∈R ∀ a∈A
So, these three (1,1),(2,2),(3,3) elements should always be present for relation to be reflexive.
Now total number elements in n(A×A)=9
Total ordered pairs =9
Among them 3 should be there for reflexive relation.
From remaining 6 elements, No of ways in which elements can be selected = all possible selections from 6 distinct elements
= 6C0+ 6C1+... 6C6=26=64
So ,64 reflexive relations can be defined on A.