Let A and B be two sets having 3 and 6 elements respectively. Write the minimum number of elements that $\left( {A \cup B} \right)$ can have.

Hint: The set of 6 elements can have all the three elements same from the smaller set containing only 3 elements, thus this leaves us with only 6 distinct elements out of a total 9 elements, 6 in one set and 3 in other. Use this concept to get the answer.

Complete step-by-step answer:
It is given that A and B be two sets having 3 and 6 elements respectively.
Let set A = $\left\{ {a,b,c} \right\}$
And set B = $\left\{ {a,b,c,d,e,f} \right\}$
As we see that set A has 3 elements and set B has 6 elements (a, b and c) are common elements.
Now he has to find out the minimum number of elements in $\left( {A \cup B} \right)$.
As we know $\left( {A \cup B} \right)$ is nothing but have all the elements in set A and B and the common elements are written only one time in the set of$\left( {A \cup B} \right)$.
$ \Rightarrow \left( {A \cup B} \right) = \left\{ {a,b,c,d,e,f} \right\}$
So this is the required set of$\left( {A \cup B} \right)$.
So $\left( {A \cup B} \right)$ can have 6 minimum number of elements.
So this is the required answer.

Note: In sets and relations $\left( {A \cup B} \right)$ implies occurrence of A event or B event, and it has a distinct formula often used which is $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ but this formula can’t be used in the above problem. This $(A \cap B)$ means occurrence of both event A as well as event B. Knowledge of the physical significance of these terms helps during sets and relations problems.