यदि \(A=\{a,b,c\}\), तो (\mathcal{P}(A)) में ठीक विषम संख्या के तत्वों वाले कितने उपसमुच्चय हैं?

If \(A=\{a,b,c\}\), how many subsets in (\mathcal{P}(A)) have an odd number of elements?

Explanation opens after your attempt
Correct Answer

A. (4)

Step 1

Concept

The odd sizes are (1) and (3), so the count is \(\binom{3}{1}+\binom{3}{3}=3+1=4\). Count subsets by their sizes.

Step 2

Why this answer is correct

The correct answer is A. (4). The odd sizes are (1) and (3), so the count is \(\binom{3}{1}+\binom{3}{3}=3+1=4\). Count subsets by their sizes.

Step 3

Exam Tip

विषम आकार (1) और (3) हैं, इसलिए संख्या \(\binom{3}{1}+\binom{3}{3}=3+1=4\) है। आकार के आधार पर उपसमुच्चय गिनें।

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Mathematics Answer, Explanation and Revision Hints

यदि \(A=\{a,b,c\}\), तो (\mathcal{P}(A)) में ठीक विषम संख्या के तत्वों वाले कितने उपसमुच्चय हैं? / If \(A=\{a,b,c\}\), how many subsets in (\mathcal{P}(A)) have an odd number of elements?

Correct Answer: A. (4). Explanation: विषम आकार (1) और (3) हैं, इसलिए संख्या \(\binom{3}{1}+\binom{3}{3}=3+1=4\) है। आकार के आधार पर उपसमुच्चय गिनें। / The odd sizes are (1) and (3), so the count is \(\binom{3}{1}+\binom{3}{3}=3+1=4\). Count subsets by their sizes.

Which concept should I revise for this Mathematics MCQ?

The odd sizes are (1) and (3), so the count is \(\binom{3}{1}+\binom{3}{3}=3+1=4\). Count subsets by their sizes.

What exam hint can help solve this Mathematics question?

विषम आकार (1) और (3) हैं, इसलिए संख्या \(\binom{3}{1}+\binom{3}{3}=3+1=4\) है। आकार के आधार पर उपसमुच्चय गिनें।