Question

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

Answer 1

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

$LHS\Rightarrow \frac{28!(2r-4)!(28-2r)!}{2r!(28-2r)!\left(24\right)!}$

$\Rightarrow \frac{28\times 27\times 26\times 25\times 24!(2r-4)!}{\left(2r\right)(2r-1)(2r-2)(2r-3)(2r-4)!\left(24\right)!}$

$\Rightarrow \frac{28\times 27\times 26\times 25}{\left(2r\right)(2r-1)(2r-2)(2r-3)}$

Put $(r=7)$

$\Rightarrow \frac{28\times 27\times 26\times 25}{14\times 13\times 12\times 11}=\frac{225}{11}$

$LHS=RHS$

Put $(r=6)$

$\Rightarrow \frac{28\times 27\times 26\times 25}{12\times 11\times 10\times 9}=\frac{455}{11}$

$\therefore$ Values of $r=7$