Question

If $A=\{x:xϵW,3\le x<6\},B=\{3,5,7\}$ and $C=\{2,4\}$;

find : $A\times (B-C)$

Find the number of elements in such a set.

Answer 1

$A=\{x:x\in W,3\le x<6\}$
$⟹A=\{3,4,5\}$
$B=\{3,5,7\}$ and
$C=\{2,4\}$
Now, $B-C=$ contains all elements present only in $B$ NOT in $C$
$B-C=\{3,5,7\}-\{2,4\}$
$B-C=\{3,5,7\}$
$A\times (B-C)=\{3,4,5\}\times \{3,5,7\}$
$=\{(3,3),(3,5),(3,7),(4,3),(4,5),(4,7),(5,3),(5,5),(5,7)\}$

Thus there are a total of $9$ elements in the above set.