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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