यदि \(A=\{0,1\}\), \(B=\{2,4\}\) और \(C=\{4,6,8\}\) हैं, तो (A\times\(B\cup C\)) में कितने अवयव होंगे?

If \(A=\{0,1\}\), \(B=\{2,4\}\) and \(C=\{4,6,8\}\), how many elements are in (A\times\(B\cup C\))?

Explanation opens after your attempt
Correct Answer

A. (8)

Step 1

Concept

\(B\cup C={2,4,6,8}\), so (n(A\times\(B\cup C\))=2\times4=8). Do not count common elements twice in a union.

Step 2

Why this answer is correct

The correct answer is A. (8). \(B\cup C={2,4,6,8}\), so (n(A\times\(B\cup C\))=2\times4=8). Do not count common elements twice in a union.

Step 3

Exam Tip

\(B\cup C={2,4,6,8}\), इसलिए (n(A\times\(B\cup C\))=2\times4=8)। संघ में समान अवयव को दो बार न गिनें।

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Mathematics Answer, Explanation and Revision Hints

यदि \(A=\{0,1\}\), \(B=\{2,4\}\) और \(C=\{4,6,8\}\) हैं, तो (A\times\(B\cup C\)) में कितने अवयव होंगे? / If \(A=\{0,1\}\), \(B=\{2,4\}\) and \(C=\{4,6,8\}\), how many elements are in (A\times\(B\cup C\))?

Correct Answer: A. (8). Explanation: \(B\cup C={2,4,6,8}\), इसलिए (n(A\times\(B\cup C\))=2\times4=8)। संघ में समान अवयव को दो बार न गिनें। / \(B\cup C={2,4,6,8}\), so (n(A\times\(B\cup C\))=2\times4=8). Do not count common elements twice in a union.

Which concept should I revise for this Mathematics MCQ?

\(B\cup C={2,4,6,8}\), so (n(A\times\(B\cup C\))=2\times4=8). Do not count common elements twice in a union.

What exam hint can help solve this Mathematics question?

\(B\cup C={2,4,6,8}\), इसलिए (n(A\times\(B\cup C\))=2\times4=8)। संघ में समान अवयव को दो बार न गिनें।