<s>[INST] Find the coefficient of $x^2y^2z^4$ in the expansion of $(x-2y+z)^8$. I need help solving this question​. How do I start?
Attempt at solution
*The term that contains $x^2y^2z^4$ is ${8 \choose 4} (-2y)^2(z)$ ?
*
 [/INST] We  can also iteratively apply the binomial theorem in order to find the coefficients. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$. 

We obtain
  \begin{align*}
[x^2y^2z^4](x-2y+z)^8&=[x^2y^2z^4]\sum_{j=0}^8\binom{8}{j}x^j(-2y+z)^{8-j}\tag{1}\\
&=[y^2z^4]\binom{8}{2}(-2y+z)^6\\
&=\binom{8}{2}[y^2z^4]\sum_{j=0}^6\binom{6}{j}(-2y)^jz^{6-j}\tag{2}\\
&=\binom{8}{2}\binom{6}{2}(-2)^2\\
&=1680
\end{align*}

Comment:


*

*In (1) we apply the binomial theorem and select the coefficient of $x^2$.

*In (2) we again apply the binomial theorem and select the coefficient  of $y^2z^4$.
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