Combinations equation solving with factorial

Question

I was trying to solve the equation using factorial as shown below but now I'm stuck at this level and need help.

$$C(n,3) = 2*C(n,2)$$

$$\frac{n!}{3!(n-3)!} = 2\frac{n!}{2!(n-2)!}$$

$$3! (n - 3)! = (n - 2)!$$

Answer 1

$$\require{cancel} \frac{n!}{3!\cdot(n-3)!} = 2\cdot\frac{n!}{2!\cdot(n-2)!}$$

Dividing by $n!$ and collecting constants:

$$\frac{1}{(n-3)!} = \frac{6}{(n-2)!}$$

$$6=\frac{(n-2)!}{(n-3)!}=\frac{(n-2)\cancel{(n-3)(n-4)(n-5)...}}{\cancel{(n-3)(n-4)(n-5)...}}$$

$$6=n-2$$

$$\therefore n=8$$

Answer 2

$$3!(n - 3)! = (n - 2)! \Leftrightarrow 6=n-2 \Leftrightarrow n=8$$