Question

If ${}^n{C}_{r-1}=36,{}^n{C}_r=84$ and ${}^n{C}_{r+1}=126,$ then find $r$.

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (B)

3

Here
$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{84}{36}$
$\Rightarrow \frac{n-r+1}{r}=\frac{7}{3}(\because \frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r})$
$\Rightarrow 3n-3r+3=7r$
$\Rightarrow 10r-3n=3\left(i\right)$
And
$\frac{{}^n{C}_{r+1}}{{}^n{C}_r}=\frac{n-(r+1)+1}{(r+1)}=\frac{126}{84}(\because \frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r})$
$\Rightarrow \frac{n-r}{r+1}=\frac{3}{2}$
$\Rightarrow 2n-2r=3r+3$
$\Rightarrow 5r-2n=-3$
$\Rightarrow 10r-4n=-6...\left(ii\right)$
Subtracting $\left(ii\right)$ from $\left(i\right)$, we get $n=9$
from $\left(i\right),10r-27=3$
$\Rightarrow 10r=30$
$\therefore r=3$.