Question

If $\frac{{}^n{C}_r}{{}^n{C}_{r+1}}=\frac{1}{2}$ and $\frac{{}^n{C}_{r+1}}{{}^n{C}_{r+2}}=\frac{2}{3}$ ; determine the values of $n$ and $r$.

Answer 1

$\frac{{}^n{\text{C}}_r}{{}^n{\text{C}}_{r+1}}=\frac{1}{2}$

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

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

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

$\Rightarrow n=3r+1$...(1)

Also,

$\frac{{}^n{\text{C}}_{r+1}}{{}^n{\text{C}}_{r+2}}=\frac{2}{3}$

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

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

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

$\Rightarrow 3r+6=2n-2r-2$

$\Rightarrow 5r=2n-8$

$\Rightarrow 5r=2(3r+1)-8$

$\Rightarrow 5r=6r+2-8$

$r=6$

$n=3r+1=3\left(6\right)+1=19$