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 .
 A: We have $9$ copies of $x-y+2z-2w$. We need to choose $3$ factors for the $x^3$, $3$ factors for the $(-y)^3$, and $1$ factor for the $(2z)$; the remaining two factors will be for the $(-2w)^2$. This produces a term $-8 x^3 y^3 z w^2$.
It remains to count how many ways to make the choices above. This is $\binom{9}{3} \binom{6}{3} \binom{3}{1} = \frac{9!}{3!3!1!2!} = 7! = 5040$. So the coefficient is $5040 \cdot (-8) = -40320$.
A: 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 $w^2$
