Question

If $n\left( A\times B \right)\cap \left( C\times D \right)=n\left[ \left( A\cap C \right)\times \left( B\cap D \right) \right]$
A.Statement-1 is true, statement-2 is true. Statement-2 is a correct explanation for statement-1.
B.Statement-1 is true, statement-2 is true. Statement-2 is not a correct explanation for statement-1.
C.Statement-1 is true, statement-2 is false.
D.Statement-1 is true, statement-2 is true.

Answer 1

Hint: We will first assume the elements of the set A and set B. Then, we will find $A\times B$ and $B\times A$ and we will check the number of common elements in the set $A\times B$ and $B\times A$. If we will get ${{9}^{2}}$ elements common, then the first statement is true. We will then check statement 2 and then its relation with statement 1.

Complete step-by-step answer:
Let the set A be$\left\{ 1,2,3,4,5,6,7,8,9,15 \right\}$ and set B be$\left\{ 1,2,3,4,5,6,7,8,9,25 \right\}$. Here set A and set B have 9 elements in common and both sets are of 10 elements.
We will first find the product of set A and set B i.e. $A\times B$
$A\times B=\left\{ 1,2,3,4,5,6,7,8,9,15 \right\}\times \left\{ 1,2,3,4,5,6,7,8,9,25 \right\}$
Thus, the product is given by:-
$A\times B=\left\{ \begin{align}
  & \left( 1,1 \right),.............,\left( 1,25 \right) \\
 & \left( 2,1 \right),\left( 2,2 \right),...........,\left( 1,25 \right) \\
 & \left( 3,1 \right),\left( 3,2 \right),..........,\left( 1,25 \right) \\
 & ............ \\
 & \left( 9,1 \right),\left( 9,2 \right),...........,\left( 1,25 \right) \\
 & \left( 15,1 \right),\left( 15,2 \right),..........,\left( 1,25 \right) \\
\end{align} \right\}$
Similarly, we will find the product
$B\times A=\left\{ \begin{align}
  & \left( 1,1 \right),.............,\left( 1,15 \right) \\
 & \left( 2,1 \right),\left( 2,2 \right),...........,\left( 1,15 \right) \\
 & \left( 3,1 \right),\left( 3,2 \right),..........,\left( 1,15 \right) \\
 & ............ \\
 & \left( 9,1 \right),\left( 9,2 \right),...........,\left( 1,15 \right) \\
 & \left( 25,1 \right),\left( 15,2 \right),..........,\left( 1,15 \right) \\
\end{align} \right\}$
We have observed that there are 81 elements common in set and, these are:-
$\left\{ \begin{align}
  & \left( 1,1 \right),.............,\left( 1,9 \right) \\
 & \left( 2,1 \right),\left( 2,2 \right),...........,\left( 1,9 \right) \\
 & \left( 3,1 \right),\left( 3,2 \right),..........,\left( 1,9 \right) \\
 & ............ \\
 & \left( 9,1 \right),\left( 9,2 \right),...........,\left( 1,9 \right) \\
\end{align} \right\}$
Hence, the statement 1 is true.
Now, we will check statement 2.
The given statement is
$n\left( A\times B \right)\cap \left( C\times D \right)=n\left[ \left( A\cap C \right)\times \left( B\cap D \right) \right]$
Yes, the statement is true. Let’s understand it with the same example.
The products of two sets are $A\times B$ and $B\times A$.
We know, number of elements common in both sets $A\times B$ and $B\times A$ are${{9}^{2}}$ which is given by $n\left( A\times B \right)\cap \left( B\times A \right)={{9}^{2}}$
We know, number of elements common in set A and set B is 9 which is given by $A\cap B$and we know the product of their intersection will have 81 elements i.e. $n\left[ \left( A\cap B \right)\times \left( B\cap A \right) \right]={{9}^{2}}$
Thus, we can write
$n\left( A\times B \right)\cap \left( B\times A \right)=n\left[ \left( A\cap B \right)\times \left( B\cap A \right) \right]$, the statement is similar to the statement 2 i.e.$n\left( A\times B \right)\cap \left( C\times D \right)=n\left[ \left( A\cap C \right)\times \left( B\cap D \right) \right]$
Hence, statement 2 is also true.
Therefore, Statement-1 is true, statement-2 is true. Statement-2 is a correct explanation for statement-1.
Thus, the correct option is A.

Note: Here we have calculated the product of two sets and also found their intersection. Let set A and set B are two sets, then the Cartesian product of set A and set B is denoted by $A\times B$ and it is defined as a set which consists of all ordered pair $\left( a,b \right)$ for which $a\in A$ and $b\in B$ and intersection of set A and set B is denoted by $A\cap B$ and it shows the common elements between the set A and set B.