Question

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

Answer 1

$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{84}{36}$

$\Rightarrow \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)!}}=\frac{84}{36}$

$\Rightarrow \frac{n-r+1}{r}=\frac{7}{3}$

$\Rightarrow 3n-3r+3=7r$

$\Rightarrow 10r=3n+3$---------------1

$\frac{{}^n{C}_{r+1}}{{}^n{C}_r}=\frac{126}{84}$

$\Rightarrow \frac{\frac{n!}{(r+1)!(n-r-1)!}}{\frac{n!}{\left(r\right)!(n-r)!}}=\frac{3}{2}$

$\Rightarrow \frac{n-r}{r+1}=\frac{3}{2}$

$\Rightarrow 2n-2r=3r+3$

$\Rightarrow 5r=2n-3$-------------------2

Dividing Equation 1 by Equation 2,

$\frac{10r}{5r}=\frac{3n+3}{2n-3}$

$\Rightarrow 2(2n-3)=3n+3$

$\Rightarrow 4n-6=3n+3$

$\Rightarrow n=6+3$

$\Rightarrow n=9$

$10r=3n+3$

$\Rightarrow 3\left(9\right)+3=30$

$\Rightarrow 10r=30$

$\Rightarrow r=3$

$\therefore n=9,r=3$