Let A and B two sets such that $n\left[ P\left( A \right) \right]=n\left[ P\left( B \right) \right]$ If (a,2), (b,3), (c,3) are in ... Find A and B where a, b, c are distinct. Find $A\times B$ ?

Hint: First at all, we should know the expression used in the above question, i.e. $n\left[ P\left( A \right) \right]\text{ and }n\left[ P\left( B \right) \right]$. It is basically related to power sets and its properties. Once we know the formula for these expressions, then we will easily identify the clue or approach for getting a solution. If we find out the number of elements in set A and set B, then by using the data given for $A\times B$ we can find set A and set B and further find $A\times B$ fully. In the question, the data given for $A\times B$ is partly why it is asking for $A\times B$ at the end of the question.

Complete step-by-step answer:
Now, let us see the given data:
\[\begin{align}
  & n\left[ P\left( A \right) \right]=8 \\
 & n\left[ P\left( B \right) \right]=4 \\
\end{align}\]
(a,2), (b,3), (c,3) are in $A\times B$
We know, if A is any set then the set of all subsets of A is called the power set of A and is denoted by P(A).
$n\left[ P\left( A \right) \right]$ is the number of elements in the power set P(A).
We know the formula for this, i.e.
\[\Rightarrow n\left[ P\left( A \right) \right]={{2}^{x}}\]
Where, x is the number of elements in set (A)
Hence, from given data,
\[\begin{align}
  & \Rightarrow n\left[ P\left( A \right) \right]={{2}^{x}}=8 \\
 & \Rightarrow {{2}^{x}}={{2}^{3}} \\
\end{align}\]
By comparing we get x = 3.
Hence, the number of elements in set A would be 3.
Similarly,
\[\begin{align}
  & \Rightarrow n\left[ P\left( B \right) \right]={{2}^{y}}=4 \\
 & \Rightarrow {{2}^{y}}={{2}^{2}} \\
\end{align}\]
By comparing we have y = 2.
Hence, the number of elements in set B would be 2.
Let,
\[\text{Set }A=\left\{ {{x}_{1}},{{x}_{2}},{{x}_{3}} \right\}\text{ and Set }B=\left\{ {{y}_{1}},{{y}_{2}} \right\}\]
Hence,
\[A\times B=\left\{ \left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{1}},{{y}_{2}} \right),\left( {{x}_{2}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{1}} \right),\left( {{x}_{3}},{{y}_{2}} \right) \right\}\cdots \cdots \left( i \right)\]
Given data,
\[\begin{matrix}
   A\times B=\left\{ \cdots \left( a,2 \right)\cdots \left( b,3 \right)\cdots \left( c,3 \right) \right\}\cdots \cdots \left( ii \right) \\
   \downarrow \\
   \text{Pairs are randomly placed} \\
\end{matrix}\]
Don’t focus on which location they are located. This is just the sample picture.
By comparing (ii) with (i), we will get:
\[\begin{align}
  & {{x}_{1}}=a \\
 & {{x}_{2}}=b \\
 & {{x}_{3}}=c \\
 & {{y}_{1}}=2 \\
 & {{y}_{2}}=3 \\
\end{align}\]
Where a, b and c are distinct.
Hence, we have
\[\text{Set A}=\left\{ a,b,c \right\}\text{ and Set B}=\left\{ 2,3 \right\}\]
Therefore,
\[\text{Set }A\times B=\left\{ \left( a,2 \right),\left( a,3 \right),\left( b,2 \right),\left( b,3 \right),\left( c,2 \right),\left( c,3 \right) \right\}\]

Note: Look in above solution, we have taken $\left( {{x}_{1}}=a \right)$ but it can also be b or c. Because in question, there is no discussion about whether (a,2), (b,3), (c,3) are present in ordered way or not in set $A\times B$
One important thing whose students should be aware in case of $A\times B$ i.e.
Cartesian product \[C=A\times B\]
Let, A = {a,b} and B = {0,1}
\[C=A\times B=\left\{ \left( a,0 \right),\left( a,1 \right),\left( b,0 \right),\left( b,1 \right) \right\}\]
We can’t change $\left( a,0 \right)\to \left( 0,a \right)$ but we can change like $\left\{ \left( a,0 \right),\left( a,1 \right),\left( b,0 \right),\left( b,1 \right) \right\}$