यदि \(U={1,2,\ldots,16}\), \(A={x:x\) (16) का भाजक है(}) और (B=A'), तो (n(\mathcal{P}(B))) कितना है?
If \(U={1,2,\ldots,16}\), \(A={x:x\) is a divisor of (16)(}), and (B=A'), what is (n(\mathcal{P}(B)))?
Explanation opens after your attempt
Correct Answer
A. (2048)
Concept
The divisors of (16) are (1,2,4,8,16), so (A') has (11) elements. Hence (n(\mathcal{P}(B))=2^{11}=2048).
Why this answer is correct
The correct answer is A. (2048). The divisors of (16) are (1,2,4,8,16), so (A') has (11) elements. Hence (n(\mathcal{P}(B))=2^{11}=2048).
Exam Tip
(16) के भाजक (1,2,4,8,16) हैं, इसलिए (A') में (11) तत्व हैं। अतः (n(\mathcal{P}(B))=2^{11}=2048)।
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