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If \(\frac{{ }^{n+1} C_{r+1}}{{ }^{n+1} C_r}=\frac{n-r+1}{m}\), then \(m=\)
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Verified Answer
The correct answer is:
\(r+1\)
\(\begin{aligned}
& \frac{{ }^{n+1} C_{r+1}}{{ }^{n+1} C_r}=\frac{n-r+1}{m} \\
& \Rightarrow \frac{\frac{(n+1) !}{(r+1) !(n-r) !}}{\frac{(n+1) !}{r !(n-r+1) !}}=\frac{n-r+1}{m} \\
& \Rightarrow \quad \frac{n-r+1}{r+1}=\frac{n-r+1}{m} \Rightarrow m=r+1
\end{aligned}\)
Hence, option (c) is correct.
& \frac{{ }^{n+1} C_{r+1}}{{ }^{n+1} C_r}=\frac{n-r+1}{m} \\
& \Rightarrow \frac{\frac{(n+1) !}{(r+1) !(n-r) !}}{\frac{(n+1) !}{r !(n-r+1) !}}=\frac{n-r+1}{m} \\
& \Rightarrow \quad \frac{n-r+1}{r+1}=\frac{n-r+1}{m} \Rightarrow m=r+1
\end{aligned}\)
Hence, option (c) is correct.
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