Question

If ${}^n{C}_3:{}^{n-1}{C}_4=20:21$, then $n$ =

Answer 1

$\frac{{}^n{C}_3}{{}^{n-1}{C}_4}=\frac{20}{21}$

$\Rightarrow \frac{n!}{3!(n-3)!}\times \frac{4!(n-5)!}{(n-1)!}=\frac{20}{21}$

$\Rightarrow \frac{n(n-1)!}{3!(n-3)(n-4)(n-5)!}\times \frac{4\times 3!(n-5)!}{(n-1)!}=\frac{20}{21}$

$\Rightarrow \left(4n\right)21=20(n^2-7n+12)$

$\Rightarrow 5n^2-56n+60=0$

$\Rightarrow n=\frac{56\pm \sqrt{{56}^2-4\left(5\right)\left(60\right)}}{10}$

$=\frac{56\pm \sqrt{3136-1200}}{10}$

$=\frac{56\pm 44}{10}=\frac{100}{10}$ or $\frac{12}{10}$

$\Rightarrow 10$ or $\frac{6}{5}$

$\Rightarrow n=10$ $(n\in Z)$.