If n(A) = 20, n(B) = 10, and if a ≤ n(A △ B) ≤ b where a, b ∈ I +, then a + b is equal to

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A

B

n (A) = 20, n (B) = 30

n (A \cup B) = 40

n (A \cap B)

n (A) = 20, n (B) = 30

n (A \cup B) = 40

n (A \cap B)

If A and B be any two sets where $n (A - B) = 8$ , $n (A \cup B) = 20$ and $n (A \cap B) = 6$ , then $n (B - A)$ is

n (A) = 50, n (B) = 20, n (A \cap B) = 10,

n (A Δ B) =

n (A \cup B)

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