Question:
If 15Cr : 15Cr − 1 = 11 : 5, find r.
Solution:
Given:
15Cr : 15Cr − 1 = 11 : 5
We have,
$\frac{{ }^{15} C_{r}}{{ }^{15} C_{r-1}}=\frac{11}{5}$
$\Rightarrow \frac{15-r+1}{r}=\frac{11}{5} \quad\left[\because \frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{n-r+1}{r}\right]$
$\Rightarrow 75-5 r+5=11 r$
$\Rightarrow 16 r=80$
$\Rightarrow r=5$