Question

Let the number of elements of the sets A and B be p and q respectively. Then the number of relations from the set A to the set B is
A.\[{2^{p + q}}\]
B.\[{2^{pq}}\]
C.\[p + q\]
D.\[pq\]

Answer 1

Hint: If A and B are two non-empty sets, then their Cartesian product A x B is the set of all ordered pairs of elements from A and B.
\[A \times B = \left\{ {\left( {x,y} \right):x \in A,y \in B} \right\}\]
To solve this question we will first calculate the number of elements in the Cartesian product of A x B which is calculated as \[n\left( {A \times B} \right) = n\left( A \right) \times n\left( B \right)\]

Complete step-by-step answer:
Given that A and B are two subsets where p is the number of elements in subset A and q is the number of elements in subset B.
This can be represented as \[n\left( A \right) = p\] and \[n\left( B \right) = q\]
Now we have to find \[A \times B\]
We know the for two sets P and Q the Cartesian product \[P \times Q\] is given as
\[P \times Q = \left\{ {\left( {p,q} \right):p \in P and q \in Q} \right\}\]
The numbers of element in \[P \times Q\] is calculated as \[n\left( {P \times Q} \right) = n\left( P \right) \times n\left( Q \right)\]
And the formula to calculate the number of subsets of \[P \times Q\] is given by \[{2^{n\left( {P \times Q} \right)}}\]
So as we already know
\[n\left( A \right) = p\]
\[n\left( B \right) = q\]
So the number of elements in \[A \times B\]will be
\[
\Rightarrow n\left( {A \times B} \right) = n\left( A \right) \times n\left( B \right) \\
   = p \times q \\
   = pq \;
 \]
Now as we know the number of subsets is given by \[{2^{n\left( {P \times Q} \right)}}\]
So the number of subset of \[A \times B\] will be
\[
\Rightarrow A \times B = {2^{n\left( {A \times B} \right)}} \\
   = {2^{pq}} \;
 \]
Therefore the number of relations from the set A to the set B is \[ = {2^{pq}}\]
So, the correct answer is “Option B”.

Note: Students must note that if the given two sets are null then the Cartesian product of those two sets will also be a null.
\[
  n\left( {A \times B} \right) = n\left( A \right) \times n\left( B \right) \\
   = \emptyset \times \emptyset \\
   = \emptyset \\
 \]