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**Hint:**For any of the random given set, let \[{\text{A = \{ a\} ,B = \{ b\} }}\]than \[{\text{A}} \times {\text{B}} = \{ (a,b)\} \] apply this concept in the above given question, and we can continue with the calculation of both the terms and we can show that \[{\text{A}} \times {\text{B}} \ne {\text{B}} \times {\text{A}}\].

**Complete step by step answer:**

As per the given sets are \[{\text{A = \{ 0,1\} }}\]and \[{\text{B = \{ 1,2,3\} }}\]

Let us first calculate the term of \[{\text{A}} \times {\text{B}}\],

As, if \[{\text{A = \{ a\} ,B = \{ b\} }}\]then \[{\text{A}} \times {\text{B}} = \{ (a,b)\} \],

So we get,

\[{\text{A}} \times {\text{B}} = \{ (0,1),(0,2),(0,3),(1,1),(1,2),(1,3)\} \]

And then calculating for \[{\text{B}} \times {\text{A}}\],

\[{\text{B}} \times {\text{A}} = \{ (1,0),(1,1),(2,0),(2,1),(3,0),(3,1)\} \]

Hence, from the above sets we can clearly interpret that \[{\text{A}} \times {\text{B}} \ne {\text{B}} \times {\text{A}}\].

Hence, proved.

**Note:**A relation between two sets is a collection of ordered pairs containing one object from each set. If the object x is from the first set and the object y is from the second set, then the objects are said to be related if the ordered pair \[{\text{(x,y)}}\] is in the relation.

1)Sets are collections of well-defined objects; relations indicate relationships between members of two sets A and B, and functions are a special type of relationship where there is exactly (or at most) one relationship for each element \[{\text{a}} \in {\text{A}}\] with an element in B.

2)Relations, Cartesian product, Relation on a Set. A relation R from X to Y is a subset of the Cartesian product \[{\text{X$\times$Y}}\]. The domain of a relation R is the set of all the first components of the ordered pairs that constitute the relation. The range of R is the set of all the second components of every ordered pair in R.

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