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**Hint:**Here, we are required to find the minimum number of elements that are present in $A \cup B$ when we are given that the set $A$ is having 3 elements and set $B$ is having 6 elements. Hence, we will consider 3 cases, when set $A$ is a subset of set $B$, when one of the elements in $A$ is not in $B$ and when nothing is common in $A$ and $B$. After observing these cases we will find the required minimum possible elements in $A \cup B$.

**Complete step-by-step answer:**

According to the question,

Set $A$ is having 3 elements and Set $B$ is having 6 elements.

Now, let us assume:

Case 1: When $A$ is a subset of $B$

Let us assume,

$A = \left\{ {a,b,c} \right\}$

and $B = \left\{ {a,b,c,d,e,f} \right\}$

Hence, $A \cup B = \left\{ {a,b,c} \right\} \cup \left\{ {a,b,c,d,e,f} \right\} = \left\{ {a,b,c,d,e,f} \right\}$

Thus, when $A$ is a subset of $B$, the number of elements in $A \cup B$ are 6.

Case 2: When one of the elements in $A$ is not in $B$

Let us assume,

$A = \left\{ {a,b,c} \right\}$

and $B = \left\{ {a,b,1,2,3,4} \right\}$

Hence, $A \cup B = \left\{ {a,b,c} \right\} \cup \left\{ {a,b,1,2,3,4} \right\} = \left\{ {a,b,c,1,2,3,4} \right\}$

Thus, when one of the elements in $A$ is not in $B$ , then the number of elements in $A \cup B$ are 7.

Case 3: When nothing is common in $A$ and $B$

Let us assume,

$A = \left\{ {a,b,c} \right\}$

and $B = \left\{ {1,2,3,4,5,6} \right\}$

Hence, $A \cup B = \left\{ {a,b,c} \right\} \cup \left\{ {1,2,3,4,5,6} \right\} = \left\{ {a,b,c,1,2,3,4,5,6} \right\}$

Thus, when nothing is common in $A$ and $B$, then the number of elements in $A \cup B$ are 9.

Clearly, in every possible case, the number of elements in $A \cup B$ is greater than or equal to the number of elements present in the set containing more elements.

**Hence, we can clearly say that the minimum number of elements that $A \cup B$ can have is 6.**

**Note:**

In mathematics, a set consists of a list of elements or numbers which are enclosed in curly brackets. A set can be written in two forms, i.e. Set-builder form or the roster form.

Set-builder form is used to represent an equation, an inequality or the numbers which have some kind of relation. This is also used to represent an infinite number of elements.

Roster form is the simpler form. In this form, we separate the numbers with the help of commas and they are enclosed again, in brackets.

Usually in a question, we are given set-builder form and to solve it further, we convert it to roster form. This makes the question easier to solve.

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