Question:
If \( \alpha = {^nC_r} \) and \( \beta = {^nC_{r-1}} \), then \( 1 + \frac{\alpha}{\beta} \) is equal to
If \( \alpha = {^nC_r} \) and \( \beta = {^nC_{r-1}} \), then \( 1 + \frac{\alpha}{\beta} \) is equal to
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Memorize ratio: \( \frac{{^nC_r}}{{^nC_{r-1}}} = \frac{n-r+1}{r} \) — very useful shortcut.
Updated On: Apr 21, 2026
- \( \frac{n+1}{r-1} \)
- \( \frac{n+1}{r} \)
- \( \frac{n-1}{1} \)
- \( \frac{n-r+1}{r} \)
- \( \frac{n+1}{r+1} \)
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The Correct Option is B
Solution and Explanation
Concept:
\[
\frac{{^nC_r}}{{^nC_{r-1}}} = \frac{n-r+1}{r}
\]
Step 1: Form ratio.
\[ \frac{\alpha}{\beta} = \frac{{^nC_r}}{{^nC_{r-1}}} = \frac{n-r+1}{r} \]
Step 2: Add 1.
\[ 1 + \frac{\alpha}{\beta} = 1 + \frac{n-r+1}{r} \] \[ = \frac{r + n - r + 1}{r} = \frac{n+1}{r} \]
Step 1: Form ratio.
\[ \frac{\alpha}{\beta} = \frac{{^nC_r}}{{^nC_{r-1}}} = \frac{n-r+1}{r} \]
Step 2: Add 1.
\[ 1 + \frac{\alpha}{\beta} = 1 + \frac{n-r+1}{r} \] \[ = \frac{r + n - r + 1}{r} = \frac{n+1}{r} \]
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