A set A has 3 elements and another set B has 6 elements. Then

Question

A set A has 3 elements and another set B has 6 elements. Then (A) 3 le n (A ∪ B) le 6  (B) 3 le n (A ∪ B) le 9 (C) 6 le n (A ∪ B) le 9  (D) 0 le n (A ∪ B) le 9

Tardigrade Admin · Accepted Answer

We have min n (A ∪ B) = max n(A), n (B) = max 3, 6 = 6 Max n (A ∪ B) = n (A) + n (B) = 9 ∴ 6 le n (A ∪ B) le 9

Q. A set A has 3 elements and another set B has 6 elements. Then

$3 \leq n (A \cup B) \leq 6$

$3 \leq n (A \cup B) \leq 9$

$6 \leq n (A \cup B) \leq 9$

$0 \leq n (A \cup B) \leq 9$

We have
min n (A $\cup$ B) = max {n(A), n (B)}
= max {3, 6} = 6
Max n (A $\cup$ B) = n (A) + n (B) = 9
$∴ 6 \leq n (A \cup B) \leq 9$

Q. A set A has 3 elements and another set B has 6 elements. Then

3≤n(A∪B)≤6

3≤n(A∪B)≤9

6≤n(A∪B)≤9

0≤n(A∪B)≤9

We have min n (A ∪ B) = max {n(A), n (B)} = max {3, 6} = 6 Max n (A ∪ B) = n (A) + n (B) = 9 ∴6≤n(A∪B)≤9

$3 \leq n (A \cup B) \leq 6$

$3 \leq n (A \cup B) \leq 9$

$6 \leq n (A \cup B) \leq 9$

$0 \leq n (A \cup B) \leq 9$

We have
min n (A $\cup$ B) = max {n(A), n (B)}
= max {3, 6} = 6
Max n (A $\cup$ B) = n (A) + n (B) = 9
$∴ 6 \leq n (A \cup B) \leq 9$