Sets A and B have 3 and 6 elements respective | vedclass

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(B) The number of elements in the union of two sets is given by the formula: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.Given $n(A) = 3$ and $n(B) = 6$

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The number of elements in the union of two sets is given by the formula: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.Given $n(A) = 3$ and $n(B) = 6$.To find the minimum number of elements in $A \cup B$,we need to maximize the number of elements in the intersection $n(A \cap B)$.The maximum possible value for $n(A \cap B)$ is the number of elements in the smaller set,which is $3$ (since $A \subseteq B$ is possible).Substituting these values: $n(A \cup B) = 3 + 6 - 3 = 6$.Therefore,the minimum number of elements in $A \cup B$ is $6$.

Sets $A$ and $B$ have $3$ and $6$ elements respectively. What can be the minimum number of elements in $A \cup B$?

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