Coefficient of binomial expansion The coefficient of $x^3$ is $4$ times the coefficient of $x^2$ in the new expansion of $(1+x)^n$. Find the value of $n$.

coefficient of $x^3$ is four times the coefficient of $x^2$ 
$$(1+x)^n=\\\binom{n}{0}1^{n}x^{0}+\binom{n}{1}1^{n-1}x^{1}+{\color{DarkBlue} {\binom{n}{2}1^{n-2}x^{2} }}+{\color{Red}{\binom{n}{3}1^{n-3}x^{3}} }+...+\binom{n}{n}1^{n-n}x^{n}\\$$so 
$$\binom{n}{3}1^{n-3}=4*\binom{n}{2}1^{n-2}\\\binom{n}{3}=4\binom{n}{2}\\\frac{n(n-1)(n-2)}{3!}=4 \frac{n(n-1)}{2!} \\\frac{n-2}{6}=\frac{4}{2}\\n-2=12\\n=14$$ note that $$n(n-1) \neq 0 \rightarrow n \neq 0 ,n \neq 1$$ because if n=1 $(1+x)^1=1+x$ 
and if n=0  then $(1+x)^0=1$ these are not acceptable 
