Answer :
i. If A = Φ, find n[P(A)]
A is a set containing some elements.
P(A) = power set = the set of all subsets of A = power set of A.
As Φ represents 0 elemets.
n(A) = cardinality of A = 0
n[ P(A) ] = cardinality of P(A)
N[P(A)] = 20 = 1
ii) If n(A) = 3, find n[P(A)]
A is a set containing some elements.
P(A) = power set = the set of all subsets of A = power set of A.
n(A) = cardinality of A = 3
n[ P(A) ] = cardinality of P(A)
N[P(A)] = 2n
where n of no of elements
N[P(A)] = 23 = 8
iii) If n[P(A)] = 512 find n(A)
A is a set containing some elements.
P(A) = power set = the set of all subsets of A = power set of A.
n(A) = cardinality of A and
n[ P(A) ] = cardinality of P(A)
N[P(A)] = 2n
where n of no of elements
so 512 = 2n
so 29 = 2n
n = 9
so n(A) = 9
iv. If n[P(A)] = 1024 find n(A)
A is a set containing some elements.
P(A) = power set = the set of all subsets of A = power set of A.
n(A) = cardinality of A and n[ P(A) ] = cardinality of P(A)
let P(A) = n
then n[ P(A) ] = 2n = 1024 = 210
so n = n(A) = 10
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