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)] = 2**

ii) If n(A) = 3, find n[P(A)]

A is a set containing some elements.

^{0}= 1ii) If n(A) = 3, find n[P(A)]

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)] = 2

^{n}

where n of no of elements

**A is a set containing some elements.**

N[P(A)] = 2

iii) If n[P(A)] = 512 find n(A)

N[P(A)] = 2

^{3}= 8iii) If n[P(A)] = 512 find n(A)

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)] = 2**

^{n}

where n of no of elements

so 512 = 2

^{n}

so 2

^{9}= 2

^{n}

n = 9

**so n(A) = 9**

iv.

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) ] = 2

^{n}= 1024 = 2

^{10}

so n = n(A) = 10

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