Question

If ${}^n{C}_{r-1}=56$, ${}^n{C}_r=28$ and ${}^n{C}_{r+1}=8$ then r is equal to

• A. 8
• B. 6
• C. 5
• D. None of these

Answer 1

The correct option is B 6

From the given information, $\frac{{}^n{C}_{r-1}}{{}^n{C}_r}=2$
$\Rightarrow \frac{n!r!(n-r)!}{(r-1)!(n-(r-1\left)\right)!.n!}=2$

$\Rightarrow \frac{r}{n+1-r}=2$

$\Rightarrow r=2n+2-2r$
$\Rightarrow 3r=2n+2$...(i)

Also, $\frac{{}^n{C}_r}{{}^n{C}_{r+1}}=\frac{7}{2}$
$\Rightarrow \frac{n!(n-(r+1\left)\right)!.(r+1)!}{n!.r!.(n-r)!}=\frac{7}{2}$

$\Rightarrow \frac{r+1}{n-r}=\frac{7}{2}$

$\Rightarrow 2r+2=7n-7r$
$\Rightarrow 9r=7n-2$ ...(ii)

$\Rightarrow n=8$
Hence,
$\Rightarrow 9r=56-2$
$\Rightarrow 9r=54$
$\Rightarrow r=6$