यदि \(U={1,2,\ldots,18}\) और \(A={x:x\) (18) का अभाज्य गुणनखंड है(}), तो (n(\mathcal{P}(A'))) कितना है?
If \(U={1,2,\ldots,18}\) and \(A={x:x\) is a prime factor of (18)(}), what is (n(\mathcal{P}(A')))?
Explanation opens after your attempt
Correct Answer
A. (65536)
Concept
The prime factors of (18) are (2) and (3), so (A) has (2) elements. (A') has (16) elements and \(2^{16}=65536\).
Why this answer is correct
The correct answer is A. (65536). The prime factors of (18) are (2) and (3), so (A) has (2) elements. (A') has (16) elements and \(2^{16}=65536\).
Exam Tip
(18) के अभाज्य गुणनखंड (2) और (3) हैं, इसलिए (A) में (2) तत्व हैं। (A') में (16) तत्व होंगे और \(2^{16}=65536\)।
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