Answer
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Hint: In this question, we are given two sets A and B, m and n elements respectively. We have to find the number of elements in $A\times B$. For this, we will first find the type of elements in $A\times B$ and then use it to find the number of elements in $A\times B$.
Complete step by step answer:
Here we are given the number of elements in set A as m and number of elements in set B as n. We need to find the number of elements in $A\times B$.
Let us suppose elements in set A are as follows:
$A=\left\{ {{a}_{1}},{{a}_{2}},{{a}_{3}},\ldots \ldots \ldots {{a}_{m}} \right\}$.
And elements in set B are as follows:
$B=\left\{ {{b}_{1}},{{b}_{2}},{{b}_{3}},\ldots \ldots \ldots {{b}_{n}} \right\}$.
We need to find elements in $A\times B$. Since, $A\times B$ always has ordered pair elements, so its elements are of the form (a,b) where $a\in A\text{ and }b\in B$.
Now, every element of A can form ordered pair with B, that is,
\[\begin{align}
& \left( {{a}_{1}},{{b}_{1}} \right),\left( {{a}_{1}},{{b}_{2}} \right)\ldots \ldots \ldots \left( {{a}_{1}},{{b}_{n}} \right) \\
& \left( {{a}_{2}},{{b}_{1}} \right),\left( {{a}_{2}},{{b}_{2}} \right)\ldots \ldots \ldots \left( {{a}_{2}},{{b}_{n}} \right) \\
& \left( {{a}_{3}},{{b}_{1}} \right),\left( {{a}_{3}},{{b}_{2}} \right)\ldots \ldots \ldots \left( {{a}_{3}},{{b}_{n}} \right) \\
& \vdots \\
& \vdots \\
& \left( {{a}_{m}},{{b}_{1}} \right),\left( {{a}_{m}},{{b}_{2}} \right)\ldots \ldots \ldots \left( {{a}_{m}},{{b}_{n}} \right) \\
\end{align}\]
Hence, all these elements will lie in $A\times B$.
Since, the number of rows are m and number of columns are n. So, the number of elements becomes $m\times n$.
Therefore, the total number of elements in $A\times B$ are $m\times n$.
Note: Students should note that, there is a huge difference between elements of $A\times B\text{ and }B\times A$. Total number of elements in any Cartesian product of two sets in the product of number of elements in each set. Here, $A\times B$ denotes the Cartesian product of A and B. While listing the elements of $A\times B$ make sure that, in the ordered pair, elements of A are written first and elements of B are written second. For $B\times A$ elements of B are written first and elements of A are written second in ordered pairs. We can also find a number of elements of $A\times A,B\times B$ which will be ${{m}^{2}}\text{ and }{{n}^{2}}$ respectively.
Complete step by step answer:
Here we are given the number of elements in set A as m and number of elements in set B as n. We need to find the number of elements in $A\times B$.
Let us suppose elements in set A are as follows:
$A=\left\{ {{a}_{1}},{{a}_{2}},{{a}_{3}},\ldots \ldots \ldots {{a}_{m}} \right\}$.
And elements in set B are as follows:
$B=\left\{ {{b}_{1}},{{b}_{2}},{{b}_{3}},\ldots \ldots \ldots {{b}_{n}} \right\}$.
We need to find elements in $A\times B$. Since, $A\times B$ always has ordered pair elements, so its elements are of the form (a,b) where $a\in A\text{ and }b\in B$.
Now, every element of A can form ordered pair with B, that is,
\[\begin{align}
& \left( {{a}_{1}},{{b}_{1}} \right),\left( {{a}_{1}},{{b}_{2}} \right)\ldots \ldots \ldots \left( {{a}_{1}},{{b}_{n}} \right) \\
& \left( {{a}_{2}},{{b}_{1}} \right),\left( {{a}_{2}},{{b}_{2}} \right)\ldots \ldots \ldots \left( {{a}_{2}},{{b}_{n}} \right) \\
& \left( {{a}_{3}},{{b}_{1}} \right),\left( {{a}_{3}},{{b}_{2}} \right)\ldots \ldots \ldots \left( {{a}_{3}},{{b}_{n}} \right) \\
& \vdots \\
& \vdots \\
& \left( {{a}_{m}},{{b}_{1}} \right),\left( {{a}_{m}},{{b}_{2}} \right)\ldots \ldots \ldots \left( {{a}_{m}},{{b}_{n}} \right) \\
\end{align}\]
Hence, all these elements will lie in $A\times B$.
Since, the number of rows are m and number of columns are n. So, the number of elements becomes $m\times n$.
Therefore, the total number of elements in $A\times B$ are $m\times n$.
Note: Students should note that, there is a huge difference between elements of $A\times B\text{ and }B\times A$. Total number of elements in any Cartesian product of two sets in the product of number of elements in each set. Here, $A\times B$ denotes the Cartesian product of A and B. While listing the elements of $A\times B$ make sure that, in the ordered pair, elements of A are written first and elements of B are written second. For $B\times A$ elements of B are written first and elements of A are written second in ordered pairs. We can also find a number of elements of $A\times A,B\times B$ which will be ${{m}^{2}}\text{ and }{{n}^{2}}$ respectively.
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