adjacent_ratio se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
5 questions tagged with adjacent_ratio.
If \(\frac{^{n}C_r}{^{n}C_{r-1}}=4\), which equation is correct?
A. (n-r+1=4r)
अनुपात \(\frac{n-r+1}{r}\) होता है। इसे (4) के बराबर रखने पर (n-r+1=4r) मिलता है। / The ratio is \(\frac{n-r+1}{r}\). Setting it equal to (4) gives (n-r+1=4r).
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What is the correct simplified form of \(\frac{^{n}C_6}{^{n}C_5}\)?
B. \(\frac{n-5}{6}\)
सामान्य अनुपात \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\) है। (r=6) रखने पर \(\frac{n-5}{6}\) मिलता है। / The general ratio is \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Putting (r=6) gives \(\frac{n-5}{6}\).
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If \(\frac{^{n}C_r}{^{n}C_{r-1}}=3\), which equation is correct?
A. (n-r+1=3r)
अनुपात \(\frac{n-r+1}{r}\) होता है। इसे (3) के बराबर रखने पर (n-r+1=3r) मिलता है। / The ratio is \(\frac{n-r+1}{r}\). Setting it equal to (3) gives (n-r+1=3r).
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What is the correct simplified value of \(\frac{^{n}C_5}{^{n}C_4}\)?
A. \(\frac{n-4}{5}\)
सामान्य अनुपात \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\) है। यहाँ (r=5) रखने पर \(\frac{n-4}{5}\) मिलता है। / The general ratio is \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Putting (r=5) gives \(\frac{n-4}{5}\).
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Which multiplier connects \(^{n}C_{r+1}\) with \(^{n}C_r\)?
B. \(\frac{n-r}{r+1}\)
फैक्टोरियल रूपों का अनुपात लेने पर \(^{n}C_{r+1}=^{n}C_r\frac{n-r}{r+1}\) मिलता है। परीक्षा में adjacent terms पर ratio method उपयोग करें। / Taking the ratio of factorial forms gives \(^{n}C_{r+1}=^{n}C_r\frac{n-r}{r+1}\). In exams use ratio method for adjacent terms.
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