Question

If ${{}^{28}C}_{2r}:{{}^{24}C}_{2r-4}=225:11$, find $r$.

Answer 1

$⟹\frac{28!}{\left(2r\right)!(28-2r)!}\times \frac{(2r-4)!(28-2r)!}{\left(24\right)!}=\frac{225}{11}$
$⟹\frac{28\times 27\times 26\times 25}{2r(2r-1)(2r-2)(2r-3)}=\frac{225}{11}$
$⟹14\times 3\times 26\times 11=2r(2r-1)(2r-2)(2r-3)$
$⟹6006=(r^2-r)({4r}^2-8r+3)$
$⟹6006={4r}^4-12r^3+11r^2-3r$
$⟹4r^4-12r^3+{11r}^2-3r-6006=0$
$r=7$ Other $r$ are complex and negative.