If $A=\left\{ 1,2,4 \right\},B=\left\{ 2,4,5 \right\},C=\left\{ 2,5 \right\}$ , then $\left( A-C \right)\times \left( B-C \right)$ is equal to
(a) $\left\{ \left( 1,4 \right) \right\}$
(b) $\left\{ \left( 1,4 \right),\left( 4,4 \right) \right\}$
(c) $\left\{ \left( 4,1 \right),\left( 4,4 \right) \right\}$
(d) None of these

Hint: To find the value of $\left( A-C \right)\times \left( B-C \right)$ , we have to first find $\left( A-C \right)$ and $\left( B-C \right)$ . When we subtract $\left( A-C \right)$ , the result will be the set of elements that results after the intersection of A and complement of C (that is, the set elements not in C). Similarly, we have to find $\left( B-C \right)$ . When we have to take the Cartesian product of $\left( A-C \right)$ and $\left( B-C \right)$ which is the set of ordered pairs of the elements in $\left( A-C \right)$ and $\left( B-C \right)$ .

Complete step by step solution:
We have to find the value of $\left( A-C \right)\times \left( B-C \right)$ . We are given with $A=\left\{ 1,2,4 \right\},B=\left\{ 2,4,5 \right\},C=\left\{ 2,5 \right\}$
Let us first find $\left( A-C \right)$ . We have to subtract C from A. The result of this subtraction will be the set of elements that are there in A but not in C. We can see that the elements that are not in C are ${{C}^{'}}=\left\{ 1,3,4 \right\}$ . The set of elements that are common in ${{C}^{'}}$ and A are $\left\{ 1,4 \right\}$ . Hence, the value of $\left( A-C \right)$ is
$\left( A-C \right)=\left\{ 1,4 \right\}$
Now, let us find $\left( B-C \right)$ . We have ${{C}^{'}}=\left\{ 1,3,4 \right\}$ . The set of elements that are common in ${{C}^{'}}$ and B are $\left\{ 4 \right\}$ . Therefore, the value of $\left( B-C \right)$ is
$\left( B-C \right)=\left\{ 4 \right\}$
Now, let us find $\left( A-C \right)\times \left( B-C \right)$ . We know that when two sets P and Q are multiplied the result will be as follows.
$P\times Q=\left\{ \left( p,q \right):p\in P,q\in Q \right\}$
Therefore, the Cartesian product $\left( A-C \right)\times \left( B-C \right)$ will be
$\left( A-C \right)\times \left( B-C \right)=\left\{ \left( 1,4 \right),\left( 4,4 \right) \right\}$

So, the correct answer is “Option b”.

Note: Students must know how to add, subtract and multiply the sets. The sum of two sets will result in the set of all elements of the two sets. We will denote this as $A\cup B$ , for two sets A and B. The subtraction of two sets, A and B, will be the intersection of A and complement of C, that is, $A\cap {{B}^{'}}$ .