Question

If ${}^n{C}_6{:}^{n-3}{C}_3=33:4$, find n.

Answer 1

$\frac{{}^n{C}_6}{{}^{n-3}{C}_3}=\frac{33}{4}$

$\Rightarrow \frac{\frac{n!}{(n-6)!6!}}{\frac{(n-3)!}{3!(n-6)!}}=\frac{33}{4}$

$\Rightarrow \frac{n!}{6!}\times \frac{3!}{(n-3)!}=\frac{33}{4}$

$\Rightarrow n(n-1)(n-2)=11\times 3\times 5\times 2\times 3=11\times 10\times 9$

$\therefore n=11$