Let A be a set of n(>0) elements. Let $N_r$ be the number of binary relations on A and let $N_f$ be the number of functions from A to A
no of binary relation = $2^{n^2}$ no of function = $n^n$ $\begin{align} \quad&&& 2^{n^2} &&& n^{n}\end{align}$ $\begin{align} &n^{2} \log (2) &&&n \log(n) &&\text{ // apply, $\log$ on both}\end{align}$ $\begin{align}&n^{2} \log (2)\quad > &n \log(n)\end{align}$
No of relation > No of function
A. Number of binary relations on set A = Nr = 2^(n^2)
B. Number of functions on set A = Nf = n^n
C. No of relation > No of function( Nr >Nf) ,because function is restricted version of relation so it is obvious.By the way, we can check by taking log at both sides.
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