Question

If $^nC_{r-1}=28,\,^nC_r =56$ and $^nC_{r+1}=70$, then $r =$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The Correct Option is C

$\frac{^{n}C_{r}}{^{n}C_{r-1}}=\frac{56}{28}$ $\Rightarrow \frac{n!}{r ! \left(n-r\right)!}\cdot\frac{\left(r-1\right)!\left(n-r+1\right)!}{n!}=2$ $\Rightarrow \frac{n-r+1}{r}=2$ $\Rightarrow n-r + 1 = 2 r$ $\Rightarrow n=3r-1 ...\left(1\right)$ Again $ \frac{^{n}C_{r+1}}{^{n}C_{r}}=\frac{70}{56}$ $\Rightarrow\frac{n!}{\left(r+1\right)!\left(n-r-1\right)!}$ $\frac{r!\left(n-r\right)!}{n!}=\frac{5}{4}$ $\Rightarrow \frac{n-r}{r+1}=\frac{5}{4}$ $\Rightarrow 4n-4r=5r+5$ $\Rightarrow 4n=9r+5$ From $\left(1\right)$ and $\left(2\right)$, we get $12r - 4 = 9r + 5$ $\Rightarrow 3r=9$ $\Rightarrow r=3$.