Question

Sets A and B have $3$ and $6$ elements respectively. What can be the minimum number of elements in $A \cup B$
$A)3$
$B)6$
$C)9$
$D)18$

Answer 1

Hint: First, from the given, we have two sets with different numbers of elements, which are set A has three elements and set B has six elements. From this, we can say that set B is the larger set.
While acting the union of these two sets, we need to find its minimum number of elements on the set of union for the A and B.

Complete step-by-step solution:
Since \[A \cup B\] means, the set of elements on both sets will occur onto the union function.
From the set theory formula for the number of elements of the union set, we have \[n(A \cup B) = n(A) + n(B) - n(A \cap B)\]
Maximum means the intersection function \[n(A \cap B)\] will have zero, \[n(A \cap B) = 0\]
Minimum means the intersection function \[n(A \cap B)\] will have a smaller size set $A \subseteq B$ which means set A and it has three elements, \[n(A \cap B) = 3\]
Since given that we need to find the minimum number of elements on the union of the two sets, hence we have \[n(A \cap B) = 3\]
Substituting this value in the number of elements value, we get \[n(A \cup B) = n(A) + n(B) - n(A \cap B) \Rightarrow n(A \cup B) = 3 + 6 - 3\] where $n(A) = 3,n(B) = 6$ from the given.
Hence, we have \[n(A \cup B) = 6\]
Therefore, the option $B)6$ is correct.

Note: Set as take the sets A and B as $B = \{ 1,2,3,4,5,6\} ,A = \{ 1,2,3\} $ then we have \[(A \cap B) = \{ 1,2,3\} \Rightarrow n(A \cap B) = 3\]
Hence, we get \[n(A \cup B) = n(A) + n(B) - n(A \cap B) \Rightarrow n(A \cup B) = 3 + 6 - 3\] and thus \[n(A \cup B) = 6\]
Which is the minimum possibility of the two sets of union functions.
This is the concept in the set theory which is to find the total number of elements of the two sets union using the given two sets the total number and intersection we can found the required.
Since if the question is about to find the intersection value, then we will apply the same formula where the only difference is \[n(A \cap B) = n(A \cup B) - n(A) - n(B)\]