If set A have $4$ elements and set B has $7$ elements. What can be the minimum number of elements in$A \cup B$?
A. $7$
B. $14$
C. $21$
D. $11$

Hint:
$A \cup B$ means union of the set A and the set B that means the elements which are in A or in B or in both A and B. For the minimum elements in $A \cup B$, maximum elements must be common in both the sets A and B.

Complete step by step solution:
Here we are given that set A has $4$ elements and set B has $7$ elements and we need to find what can be the minimum number of elements in$A \cup B$. So we need to understand the meaning of$A \cup B$ and here we must know that the sign between A and B represents the sign of union which means that $A \cup B$ means union of the set A and the set B that means the elements which are in A or in B or in both A and B.
Let us understand it by the simple example:
If we have the sets A and B as
$
\Rightarrow A = \{ 1,2,4\} \\
\Rightarrow B = \{ 1,5,3,6\} \\
\Rightarrow A \cup B = \{ 1,2,3,4,5,6\} \\
 $
So we get that in the$A \cup B$, we have put both the elements which were contained in the sets A and B but as $1$ was in both the sets so we counted it only once and not twice.
So for the minimum number of elements in $A \cup B$ we need to have the maximum common elements as it is understood from the example and also from the formula which says.
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
So as we know that we are given that set A has $4$ elements and set B has $7$ elements and we need to find what can be the minimum number of elements in$A \cup B$.
So as out of set A and B we have minimum elements in the set A which is $4$ so the case is possible if all those four elements come in the common intersection of the set A and set B which means that
$
\Rightarrow n(A) = 4 \\
\Rightarrow n(B) = 7 \\
\Rightarrow n(A \cap B) = 4 \\
 $
So we can get that $n(A \cup B) = n(A) + n(B) - n(A \cap B)$

$n(A \cup B) = 4 + 7 - 4 = 7$

Note:
Similarly for finding the maximum value in the set theory of the number of elements in $(A \cup B)$ we need to know that $n(A \cap B) = 0$
Hence we will get that $n(A \cup B) = n(A) + n(B) - n(A \cap B)$
$n(A \cup B) = 4 + 7 - 0 = 11$