यदि \(^{n}C_{r+1}=\frac{n-r}{r+1},^{n}C_r\) है तो यह संबंध किस ratio से निकला है?

If \(^{n}C_{r+1}=\frac{n-r}{r+1},^{n}C_r\) then this relation is derived from which ratio?

Explanation opens after your attempt
Correct Answer

B. \(\frac{^{n}C_{r+1}}{^{n}C_r}=\frac{n-r}{r+1}\)

Step 1

Concept

Writing factorial forms and canceling common terms gives the ratio. In exams ratio method is fast for consecutive combinations.

Step 2

Why this answer is correct

The correct answer is B. \(\frac{^{n}C_{r+1}}{^{n}C_r}=\frac{n-r}{r+1}\). Writing factorial forms and canceling common terms gives the ratio. In exams ratio method is fast for consecutive combinations.

Step 3

Exam Tip

Factorial form लिखकर common terms cancel करने से ratio मिलता है। परीक्षा में consecutive combinations में ratio method तेज होता है।

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यदि \(^{n}C_{r+1}=\frac{n-r}{r+1},^{n}C_r\) है तो यह संबंध किस ratio से निकला है? / If \(^{n}C_{r+1}=\frac{n-r}{r+1},^{n}C_r\) then this relation is derived from which ratio?

Correct Answer: B. \(\frac{^{n}C_{r+1}}{^{n}C_r}=\frac{n-r}{r+1}\). Explanation: Factorial form लिखकर common terms cancel करने से ratio मिलता है। परीक्षा में consecutive combinations में ratio method तेज होता है। / Writing factorial forms and canceling common terms gives the ratio. In exams ratio method is fast for consecutive combinations.

Which concept should I revise for this Mathematics MCQ?

Writing factorial forms and canceling common terms gives the ratio. In exams ratio method is fast for consecutive combinations.

What exam hint can help solve this Mathematics question?

Factorial form लिखकर common terms cancel करने से ratio मिलता है। परीक्षा में consecutive combinations में ratio method तेज होता है।