<s>[INST] Find the co-efficient of $x^3y^3zw^2$ in the expression of $(x-y+2z-2w)^9$. 
Find the co-efficient of $x^3y^3zw^2$ in the expression of $(x-y+2z-2w)^9$.

I have clearly understood how to use the bin theorem for $(a+b)^n$, but am unable to extend the theorem to $4$ terms .
 [/INST] It is also possible to apply the binomial theorem    iteratively. We focus on one term and keep the other terms together. This way we reduce the number of terms step by step by one, till we have only two left at the end.
To keep the calculations manageable it is convenient to use the coefficient of operator $[x^j]$ to select the coefficient of $x^j$ of a polynomial or series. This way we can write e.g.
\begin{align*}
[x^j](x+y+z)^n&=[x^j]\sum_{k=0}^n\binom{n}{k}x^k(y+z)^{n-k}\\
&=\binom{n}{j}(y+z)^{n-j}\\
\end{align*}

Using this technique we obtain
  \begin{align*}
[x^3y^3zw^2]&(x-y+2z-2w)^9\\
&=[x^3y^3zw^2]\sum_{k=0}^9\binom{9}{k}x^k(-y+2z-2w)^{9-k}\tag{1}\\
&=\binom{9}{3}[y^3zw^2](-y+2z-2w)^6\tag{2}\\
&=\binom{9}{3}[y^3zw^2]\sum_{k=0}^6\binom{6}{k}(-y)^k(2z-2w)^{6-k}\tag{3}\\
&=\binom{9}{3}\binom{6}{3}(-1)^3[zw^2](2z-2w)^3\tag{4}\\
&=-\binom{9}{3}\binom{6}{3}[zw^2]\sum_{k=0}^3\binom{3}{k}(2z)^k(-2w)^{3-k}\tag{5}\\
&=-\binom{9}{3}\binom{6}{3}\binom{3}{1}2^1(-2)^2\tag{6}\\
&=-84\cdot20\cdot3\cdot8\\
&=-40320
\end{align*}

Comment:


*

*In  (1), (3) and (5)  we apply the binomial theorem

*In (2), (4) and (6) we select the coefficient of $x^3,y^3,z$ and $w^2$
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