Let A and B be two sets such that $n\left( A \right) = 3$ and $n\left( B \right) = 2$. If $\left( {x,{\text{ }}1} \right), \left( {y,{\text{ }}2} \right), \left( {z,{\text{ }}1} \right)$ are in $A \times B$ find A and B where x, y and, z are distinct elements.

Hint: Using the roster definition of the Cartesian product of two sets from $(P \to Q)$ i.e. $(P \times Q)$ is defined as a relation, $(P \to Q)$ where the elements of $(P \times Q)$ will be in the form (p,q) where
$p \in P$ and $q \in Q$.
$P \times Q = \{ (p,q):p \in P,q \in Q\} $
We’ll find the elements of A and B, and hence find the complete set A and set B.

Complete step by step solution: Given data: $n\left( A \right) = 3$ and $n\left( B \right) = 2$
And $\left( {x,{\text{ }}1} \right), \left( {y,{\text{ }}2} \right), \left( {z,{\text{ }}1} \right)$ are in $A \times B$
Now we know that if we have two sets let P and Q then
$(P \times Q)$is defined as a relation, $(P \to Q)$ where the elements of $(P \times Q)$ will be in the form (p,q) where
$p \in P$ and $q \in Q$.
$P \times Q = \{ (p,q): p \in P,q \in Q\} $
Using the above equation we can say that
$x,y,z \in A$
And $1,2 \in B$

Therefore, $A = \{ x,y,z\} $
And $B = \{ 1,2\} $

Note: Most of the students think that in $\left( {x,{\text{ }}1} \right),\left( {y,{\text{ }}2} \right),\left( {z,{\text{ }}1} \right) \in A \times B$, element 1 is in two pair so the occurrence of 1 is will two times i.e. $B = \{ 1,1,2\} $ but it is wrong as it is given that $n\left( B \right) = 2$ and B contains only two elements and the occurrence of 1 is two times as 1 is the image of two elements of A but is not times in the set B.