Question

Find $r$, if ${{}^{10}C}_{r+2}$ : ${{}^{10}C}_r=10:21$.

Answer 1

$\frac{\frac{10!}{(r+2)!(10-r-2)!}}{\frac{10!}{r!(10-r)!}}=\frac{10}{21}$

$\Rightarrow \frac{r!\times (10-r)(9-r)\times (8-r)!}{(r+2)(r+1)r!\times (8-r)!}=\frac{10}{21}$

$\Rightarrow (10-r)(9-r)\times 21=(r+2)(r+1)\times 10$

$\Rightarrow 21(r^2-19r+90)=10(r^2+3r+2)$

$\Rightarrow 11r^2-429r+1870=0$

$r=34,5$

$\therefore r=5$