सूत्र (^{n}C_r=\frac{n!}{r!(n-r)!}) किस विचार से प्राप्त होता है?

Which idea gives (^{n}C_r=\frac{n!}{r!(n-r)!})?

Explanation opens after your attempt
Correct Answer

B. \(^{n}P_r\) को (r!) से भाग देनाDividing \(^{n}P_r\) by (r!)

Step 1

Concept

In combination extra orders are removed from permutation. In exams write \(^{n}P_r\) and divide by (r!).

Step 2

Why this answer is correct

The correct answer is B. \(^{n}P_r\) को (r!) से भाग देना / Dividing \(^{n}P_r\) by (r!). In combination extra orders are removed from permutation. In exams write \(^{n}P_r\) and divide by (r!).

Step 3

Exam Tip

Combination में permutation से अतिरिक्त क्रम हटाए जाते हैं। परीक्षा में \(^{n}P_r\) लिखकर (r!) से भाग दें।

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Mathematics Answer, Explanation and Revision Hints

सूत्र (^{n}C_r=\frac{n!}{r!(n-r)!}) किस विचार से प्राप्त होता है? / Which idea gives (^{n}C_r=\frac{n!}{r!(n-r)!})?

Correct Answer: B. \(^{n}P_r\) को (r!) से भाग देना / Dividing \(^{n}P_r\) by (r!). Explanation: Combination में permutation से अतिरिक्त क्रम हटाए जाते हैं। परीक्षा में \(^{n}P_r\) लिखकर (r!) से भाग दें। / In combination extra orders are removed from permutation. In exams write \(^{n}P_r\) and divide by (r!).

Which concept should I revise for this Mathematics MCQ?

In combination extra orders are removed from permutation. In exams write \(^{n}P_r\) and divide by (r!).

What exam hint can help solve this Mathematics question?

Combination में permutation से अतिरिक्त क्रम हटाए जाते हैं। परीक्षा में \(^{n}P_r\) लिखकर (r!) से भाग दें।