Question

If a set $A$ has ${}^{\prime }{n}^{\prime }$ distinct elements, then the number of all relations on $A$ is

• A. $2^{n^2}$
• B. $n^2$
• C. $2^n$
• D. $2^{n^n}$

Answer 1

The correct option is A $2^{n^2}$

Let $A$ and $B$ are two sets such that $n\left(A\right)=a$ and $n\left(B\right)=b.$
Then, number of relations from $A$ to $B$ is $=2^{n\left(A\right)\times n\left(B\right)}=2^{ab}$

Hence, number of relations from $A$ to $A$ is $2^{n\left(A\right)\times n\left(A\right)}=2^{n^2}$