Question

If ${}^n{C}_{r-1}=10{,}^n{C}_r=45$ and ${}^n{C}_{r+1}=120$ then $r$ equals

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

We know that

$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}$

$\therefore \frac{n-r+1}{r}=\frac{45}{10}$

$\Rightarrow \frac{n-r+1}{r}=\frac{9}{2}$

$\Rightarrow \frac{n-r+1}{r}=\frac{9}{2}$

$\Rightarrow 11r-2n=\mathrm{2............}\left(1\right)$

$\frac{{}^n{C}_{r+1}}{{}^n{C}_r}=\frac{n-(r+1)+1}{r+1}=\frac{120}{45}$

$\therefore \frac{n-r}{r+1}=\frac{8}{3}$

$\Rightarrow 11r-3n=-\mathrm{8............}\left(2\right)$

On solving (1) and (2) we get,

$n=10$ and $r=2$