यदि \(A=\{a,b,c,d,e\}\), तो (\mathcal{P}(A)) के कितने तत्वों में (a) होगा लेकिन (b) और (c) दोनों नहीं होंगे?
If \(A=\{a,b,c,d,e\}\), how many elements of (\mathcal{P}(A)) contain (a) but do not contain both (b) and (c)?
Explanation opens after your attempt
Correct Answer
A. (12)
Concept
(a) is fixed and (d,e) are free. For (b,c), remove the one case where both are chosen, so the count is \(3\times 2^2=12\).
Why this answer is correct
The correct answer is A. (12). (a) is fixed and (d,e) are free. For (b,c), remove the one case where both are chosen, so the count is \(3\times 2^2=12\).
Exam Tip
(a) निश्चित है और (d,e) स्वतंत्र हैं। (b,c) के लिए कुल (4) विकल्पों में से दोनों साथ वाला विकल्प हटेगा, इसलिए \(3\times 2^2=12\)।
Login to save your score, XP, coins and progress.