If ${ }^{15} C_{r}:{ }^{15} C_{r-1}=11: 5$, find $r$.
Given: ${ }^{15} \mathrm{C}_{r}:{ }^{15} \mathrm{C}_{r-1}=11: 5$
To find: $r=?$
${ }^{15} \mathrm{C}_{\mathrm{r}}:{ }^{15} \mathrm{C}_{\mathrm{r}-1}=11: 5$
$\frac{\left(\begin{array}{c}15 \\ r\end{array}\right)}{\left(\begin{array}{c}15 \\ r-1\end{array}\right)}=\frac{11}{5}$
$\frac{\frac{15 !}{(15-r) ! r !}}{\frac{15 !}{(16-r) !(r-1) !}}=\frac{11}{5}$
$\Rightarrow \frac{15 !(16-r) !(r-1) !}{(15-r) ! r ! 15 !}=\frac{11}{5}$
$\frac{(16-r) \times(15-r) ! \times(r-1) !}{(15-r) ! \times r \times(r-1) !}=\frac{11}{5}$
$\Rightarrow \frac{16-r}{r}=\frac{11}{5}$
$\Rightarrow 5 \times(16-r)=11 r$
$\Rightarrow 80-5 r=11 r$
$\Rightarrow 16 r=80$
$\Rightarrow r=\frac{80}{16}$
$\Rightarrow \mathrm{r}=5$
Ans: $r=5$