Question

Find the value of
If ${}^{15}{C}_{3r}{:}^{15}{C}_{r+1}=11:3$ find $r$.

Answer 1

Given:
$\frac{{}^{15}{C}_{3r}}{{}^{15}{C}_{r+1}}=\frac{\frac{\left(15\right)!}{\left(3r\right)!(15-3r)!}}{\frac{\left(15\right)!}{(r+1)!(15-r-1)!}}=\frac{11}{3}$

We get
$\frac{(r+1)!(15-r-1)!}{\left(3r\right)!(15-3r)!}=\frac{11}{3}$

When $r=1$
$\frac{\left(2\right)!(15-1-1)!}{\left(3\right)!(15-3)!}=\frac{13}{3}$
When $r=2$
$\frac{\left(3\right)!\left(12\right)!}{\left(6\right)!\left(9\right)!}=11$
When $r=3$
$\frac{\left(4\right)!\left(11\right)!}{\left(9\right)!\left(6\right)!}=\frac{11}{3}$
So $r=3$ is the answer.