यदि \(^{n}C_{r-2}:{}^{n}C_{r-1}:{}^{n}C_r\) दिए हों, तो formula derivation में कौन-सा ratio सबसे पहले उपयोगी होगा?

If \(^{n}C_{r-2}:{}^{n}C_{r-1}:{}^{n}C_r\) is given, which ratio is most useful first in the derivation?

Explanation opens after your attempt
Correct Answer

A. \(\frac{{}^{n}C_{r-1}}{{}^{n}C_{r-2}}=\frac{n-r+2}{r-1}\)

Step 1

Concept

Factorial cancellation in consecutive combinations gives this ratio. In exams use adjacent ratios instead of calculating long values.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{{}^{n}C_{r-1}}{{}^{n}C_{r-2}}=\frac{n-r+2}{r-1}\). Factorial cancellation in consecutive combinations gives this ratio. In exams use adjacent ratios instead of calculating long values.

Step 3

Exam Tip

Consecutive combinations में factorial cancellation से यह ratio मिलता है। परीक्षा में लंबी values निकालने के बजाय adjacent ratio लगाएं।

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यदि \(^{n}C_{r-2}:{}^{n}C_{r-1}:{}^{n}C_r\) दिए हों, तो formula derivation में कौन-सा ratio सबसे पहले उपयोगी होगा? / If \(^{n}C_{r-2}:{}^{n}C_{r-1}:{}^{n}C_r\) is given, which ratio is most useful first in the derivation?

Correct Answer: A. \(\frac{{}^{n}C_{r-1}}{{}^{n}C_{r-2}}=\frac{n-r+2}{r-1}\). Explanation: Consecutive combinations में factorial cancellation से यह ratio मिलता है। परीक्षा में लंबी values निकालने के बजाय adjacent ratio लगाएं। / Factorial cancellation in consecutive combinations gives this ratio. In exams use adjacent ratios instead of calculating long values.

Which concept should I revise for this Mathematics MCQ?

Factorial cancellation in consecutive combinations gives this ratio. In exams use adjacent ratios instead of calculating long values.

What exam hint can help solve this Mathematics question?

Consecutive combinations में factorial cancellation से यह ratio मिलता है। परीक्षा में लंबी values निकालने के बजाय adjacent ratio लगाएं।