Question

If n(P(A)) = 64 and n(P(B)) = 32, then the maximum value of n(A Δ B) is

Answer 1

Answer:

heres your answer is...

Answer 2

Solution :-

we know that,

• n(A) = Total number of elements in set A .
• P(A) = set of all subsets of set A .
• Number of subsets of power set = 2^n .

so,

→ n(P(A)) = 64

→ 2^n = 64

→ 2^n = 2^6

using a^m = a^n => m = n ,

→ n = 6

then,

→ n(A) = 6

similarly,

→ n(P(B)) = 32

→ 2^n = 32

→ 2^n = 2^5

using a^m = a^n => m = n ,

→ n = 5

then,

→ n(B) = 5

now,

→ n(A ∆ B) = n(A U B) - n(A ∩ B)

→ n(A ∆ B) = [n(A) + n(B) - n(A ∩ B)] - n(A ∩ B)

→ n(A ∆ B) = n(A) + n(B) - 2 * n(A ∩ B)

we can conclude that, for maximum value of n(A Δ B) , n(A ∩ B) must be zero .

therefore,

→ Maximum value of n(A ∆ B) = 6 + 5 = 11 (Ans.)

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