यदि \(U={1,2,\ldots,72}\), \(A={x:x\in U,4\mid x}\), \(B={x:x\in U,6\mid x}\) और \(C={x:x\in U,9\mid x}\), तो (n(\(A\cup B\cup C\)')) क्या है?

If \(U={1,2,\ldots,72}\), \(A={x:x\in U,4\mid x}\), \(B={x:x\in U,6\mid x}\), and \(C={x:x\in U,9\mid x}\), what is (n(\(A\cup B\cup C\)'))?

Explanation opens after your attempt
Correct Answer

A. (42)

Step 1

Concept

By inclusion-exclusion, (n\(A\cup B\cup C\)=18+12+8-6-2-4+2=28). Therefore the complement is (72-28=44).

Step 2

Why this answer is correct

The correct answer is A. (42). By inclusion-exclusion, (n\(A\cup B\cup C\)=18+12+8-6-2-4+2=28). Therefore the complement is (72-28=44).

Step 3

Exam Tip

समावेशन-बहिष्करण से (n\(A\cup B\cup C\)=18+12+8-6-2-4+2=28) है। इसलिए पूरक (72-28=44) नहीं बल्कि (44) है।

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Mathematics Answer, Explanation and Revision Hints

यदि \(U={1,2,\ldots,72}\), \(A={x:x\in U,4\mid x}\), \(B={x:x\in U,6\mid x}\) और \(C={x:x\in U,9\mid x}\), तो (n(\(A\cup B\cup C\)')) क्या है? / If \(U={1,2,\ldots,72}\), \(A={x:x\in U,4\mid x}\), \(B={x:x\in U,6\mid x}\), and \(C={x:x\in U,9\mid x}\), what is (n(\(A\cup B\cup C\)'))?

Correct Answer: A. (42). Explanation: समावेशन-बहिष्करण से (n\(A\cup B\cup C\)=18+12+8-6-2-4+2=28) है। इसलिए पूरक (72-28=44) नहीं बल्कि (44) है। / By inclusion-exclusion, (n\(A\cup B\cup C\)=18+12+8-6-2-4+2=28). Therefore the complement is (72-28=44).

Which concept should I revise for this Mathematics MCQ?

By inclusion-exclusion, (n\(A\cup B\cup C\)=18+12+8-6-2-4+2=28). Therefore the complement is (72-28=44).

What exam hint can help solve this Mathematics question?

समावेशन-बहिष्करण से (n\(A\cup B\cup C\)=18+12+8-6-2-4+2=28) है। इसलिए पूरक (72-28=44) नहीं बल्कि (44) है।