Question

Find the number of elements in $A\times \left( B-C \right)$, If $A=\left\{ x:x\in W,3\le x\le 6 \right\},B=\left\{ 3,5,7 \right\}$ and $C=\left\{ 2,4 \right\};$

Answer 1

Hint: The set difference gives a set which contains the exclusive elements in the prior set. The Cartesian product of two sets gives a total number of elements equal to the product of the number of elements in both sets.

Complete step-by-step answer:
We have been given the following sets:
$A=\left\{ x:x\in W,3\le x\le 6 \right\}$

Here W represents Whole numbers. So, the elements of set A will be 3, 4, 5, 6 and it will be represented as,
A = {3, 4, 5, 6}, B = {3, 5, 7}, C = {2, 4}

Substituting the values of the given sets A, B and C in the expression we have,
$A\times \left( B-C \right)=\left\{ 3,4,5,6 \right\}\times \left[ \left\{ 3,5,7 \right\}-\left\{ 2,4 \right\} \right].................(i)$

Using the BODMAS priority of the mathematical operators we need to first solve the subtraction in the brackets then we will do the multiplication on equation (i).

The set difference gives a set which contains the exclusive elements in the prior set. So, here we have 3, 5, 7 as the exclusive elements in set B which are not in set C. The equation (i) reduces to
$A\times \left( B-C \right)=\left\{ 3,4,5,6 \right\}\times \left\{ 3,5,7 \right\}$

The Cartesian product of two sets gives a total number of elements equal to the product of the number of elements in both sets. So, we will be having a total of $4\times 3=12$ elements as shown below:
$A\times \left( B-C \right)=\left\{ \left( 3,3 \right),(3,5),(3,7),(4,3),(4,5),(4,7),(5,3),(5,5),(5,7),(6,3),(6,5),(6,7) \right\}$

Hence, the total number of elements in such a set will be 12.

Note: The chances of making mistakes are when we try to subtract the set values instead of keeping the exclusive elements we take 0, as there are no common elements in the set. But this is wrong and we will get the wrong answer.