Question

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

Answer 1

$\frac{\frac{28!}{\left(2r\right)!(28-2r)!}}{\frac{24!}{(2r-4)!(24-(2r-4\left)\right)!}}=\frac{225}{11}\Rightarrow \frac{28\times 27\times 26\times 25\times 24!(2r-4)!(28-2r)!}{\left(2r\right)!(28-2r)!24!}=\frac{225}{11}\Rightarrow \frac{28\times 27\times 26\times 25}{2r\times (2r-1)\times (2r-2)(2r-3)}=\frac{225}{11}\Rightarrow \frac{28\times 27\times 26\times 25\times 11}{15\times 15}=2r(2r-1)(2r-2)(2r-3)$
$\Rightarrow 11\times 12\times 13\times 14=2r(2r-1)(2r-2)(2r-3)$
Composing both sides r = 7.