Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Using $\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}$
$\Rightarrow \frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{45}{10}$ and $\frac{{}^n{C}_{r+1}}{{}^n{C}_r}=\frac{120}{45}$
$\Rightarrow \frac{n-r+1}{r}=\frac{9}{2}$
and $\frac{n-r}{r+1}=\frac{8}{3}$
$\Rightarrow \frac{8}{3}(r+1)+1=\frac{9}{2}r$
$\Rightarrow 16r+16+6=27r$
$\Rightarrow 11r=22$
$\therefore r=2$