\({}^{n}C_r\) को (\frac{n(n-1)\cdots(n-r+1)}{r!}) लिखना किस connection को दिखाता है?

Writing \({}^{n}C_r\) as (\frac{n(n-1)\cdots(n-r+1)}{r!}) shows which connection?

Explanation opens after your attempt
Correct Answer

A. Permutation count को (r!) से divide करके combination मिलता हैCombination is obtained by dividing permutation count by (r!)

Step 1

Concept

The numerator is \(^{n}P_r\) and the denominator removes order. In exams use this form for quick numerical cancellation.

Step 2

Why this answer is correct

The correct answer is A. Permutation count को (r!) से divide करके combination मिलता है / Combination is obtained by dividing permutation count by (r!). The numerator is \(^{n}P_r\) and the denominator removes order. In exams use this form for quick numerical cancellation.

Step 3

Exam Tip

Numerator \(^{n}P_r\) है और denominator order हटाता है। परीक्षा में इस form से numerical cancellation जल्दी करें।

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\({}^{n}C_r\) को (\frac{n(n-1)\cdots(n-r+1)}{r!}) लिखना किस connection को दिखाता है? / Writing \({}^{n}C_r\) as (\frac{n(n-1)\cdots(n-r+1)}{r!}) shows which connection?

Correct Answer: A. Permutation count को (r!) से divide करके combination मिलता है / Combination is obtained by dividing permutation count by (r!). Explanation: Numerator \(^{n}P_r\) है और denominator order हटाता है। परीक्षा में इस form से numerical cancellation जल्दी करें। / The numerator is \(^{n}P_r\) and the denominator removes order. In exams use this form for quick numerical cancellation.

Which concept should I revise for this Mathematics MCQ?

The numerator is \(^{n}P_r\) and the denominator removes order. In exams use this form for quick numerical cancellation.

What exam hint can help solve this Mathematics question?

Numerator \(^{n}P_r\) है और denominator order हटाता है। परीक्षा में इस form से numerical cancellation जल्दी करें।