Very efficient coefficient

$\text {From a Property of Binomial Thm.} \\ \text{Coefficient of a term in } (a+b+c+d+...)^n \text { is :} \\ \dfrac{n!}{{n_1!}\times{n_2!}\times{n_3!}\times{n_4!}\times{....}}\times{a^{n_1}}\times{b^{n_2}}\times{c^{n_3}}\times{d^{n_4}\times{....}} \\ \text{Where } n_1+n_2+n_3+n_4+\cdots=n \text{ and } 0\le{n_1,n_2,n_3,n_4,...}\le{n}\\ \text{In this case , we got } (x+y+z)^{14} \text{ And we need to find coefficient of } x^2y^4z^8 \\ \text{coefficient of } x^2y^4z^8=\dfrac{14!}{2!\times{4!}\times{8!}} \\ \implies \dfrac{14\times{13}\times{12}\times{11}\times{10}\times{9}}{2\times{4}\times{3}\times{2}}\\ \text{Ans }=\boxed{45045}$