Total number of binary relation from n element set to itself is $2^{n^{2}}$ I.e.n^{2 }entries with two choice take it or not.
∣A∣ =n
∣A⨉A∣ = n^2
Relation is the subsets of A⨉A, so the total no of binary relation on set A = cardinality of power set of (A⨉A) = ∣ p(A⨉A) ∣ =2^(n^2)
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This might help..