Question

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

Answer 1

$\frac{{}^{28}{C}_{2r}}{{}^{24}{C}_{2r-4}}=\frac{225}{11}$

$\Rightarrow \frac{\frac{28!}{2r!(28-2r)!}}{\frac{24!}{(2r-4)!(28-2r)!}}=\frac{225}{11}$

$\Rightarrow \frac{28\times 24\times 26\times 25}{2r\times 2r-1\times 2r-2\times 2r-3}=\frac{225}{11}$

$\Rightarrow 9\left(2r\right)(2r-1)(2r-2)(2r-3)=14\times 2\times 9\times 3\times 13\times 2\times 11$

$\Rightarrow \left(2r\right)(2r-1)(2r-2)(2r-3)=14\times 13\times 12\times 11$

$\Rightarrow 2r=14$

$\Rightarrow r=7$