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If $n(A)=p$ and $n(B)=q$, then the numbers of relations from the set $A$ to the set $B$ is
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The correct answer is:
$2^{p q}$
Given, $n(A)=p$ and $n(B)=q$
$\therefore n(A \times B)=p q$
The number of relations from a set $A$ to a set $B$ is same as the total number of subset of the set $A \times B$. We know that if $n(A)=k$, then $n(P(A))=2^k$
Now, the total number of subset of $A \times B$ be $2^{p q}$
$\therefore$ Then number of relations from the set $A$ to the set $B$ is $2^{p q}$.
$\therefore n(A \times B)=p q$
The number of relations from a set $A$ to a set $B$ is same as the total number of subset of the set $A \times B$. We know that if $n(A)=k$, then $n(P(A))=2^k$
Now, the total number of subset of $A \times B$ be $2^{p q}$
$\therefore$ Then number of relations from the set $A$ to the set $B$ is $2^{p q}$.
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