Question

If ${}^{20}{C}_{r+2}{=}^{20}{C}_{2r-3}$, find ${}^{12}{C}_r$.

Answer 1

$r+2=20-2r+3$

$\Rightarrow r+2+2r-3=20$ since ${}^n{C}_x{=}^n{C}_y\Rightarrow x=n-y$

$\Rightarrow 3r-1=20$

$\Rightarrow 3r=21$ or $r=\frac{21}{3}=7$

${\therefore }^{12}{C}_7=\frac{12!}{5!7!}$

$=\frac{12\times 11\times 10\times 9\times 8\times 7!}{5\times 4\times 3\times 2\times 7!}$

$=792$