A. Consider a $2D$ matrix of $n\times n$ size where each dimension represent the $n$ elements of $A.$ So, we get $n^2$ elements where each one can be viewed as an ordered pair $(i,j), 1\leq i,j \leq n.$ Now if we consider a set with these $n^2$ elements, any subset of it will be a binary relation from $A$ to $A$.
So, no. of binary relations, $N_r = 2^{n^2}$
For a function from a set of $n$ elements to a set of $m$ elements we have $m$ choice for each of the $n$ elements of the domain. Thus we get $m^n$ possibilities. Here, $m = n.$ So,
No. of functions, $N_f = n^n$
Now, number of functions must be smaller than the number of relations because every function is also a relation. We can prove this formally as follows:
$\lim_{n\to \infty} \dfrac{2^{n^2}}{n^n} $
$\dfrac{\infty}{\infty} \text{ form}.$ So, applying L'Hopital's rule (Limit remains same after taking derivative of both numerator and denominator)
$= \lim_{n \to \infty} \dfrac{2n.\log 2. 2^{n^2}}{nn^{n-1}}$
We can continue like this applying L'Hopital's rule and we should eventually get $1$ in the denominator (as after each derivation power of $n$ decreases by $1$) which gives the limit value as $\infty.$ So, $2^{n^2}$ is faster growing than $n^n.$
So, for sufficiently large $n,$ $N_r > N_f$