Determine the coefficient of $wx^3y^2z^2$ in $(2w -x + y -2z)^8$

Question

They provide a similar example: Similarly provided example

I tried to set mine up the same way, so I had My Answer so far:

Can someone let me know if I'm even close? In the example, I have no idea no idea how the would have determined that the "6" should be squared and I'm not sure where the "1!" term came from on the denominator of their first calculation. Any tips or pointers are appreciated (please forgive my stupidity, it's been about 10 years since I took a high-school math course!).

Answer 1

Hint:

We can re-write as $$(2w-x+y-2z)^8=\sum_{r=0}^8\binom8r(2w-x)^{8-r}(y-2z)^r$$

As the sum of coefficients of $w,x$ is $1+3=4,$ we need $8-r=4\iff r=?$

Now the general term of $(2w-x)^4$ is $$\binom4k(2w)^{4-k}(-x)^k$$

We need $k=3$

Answer 2

Yes, your answer is correct: $$\frac{8!}{3!2!2!1!}\cdot (2)^1\cdot (-1)^3\cdot1^2\cdot (-2)^2=1680\cdot -8=-13440.$$ At the denominator, $1!$ comes from the exponent of $w=w^1$.

Another way: $$[wy^2x^3z^2] (2w +(y-x -2z))^8=[y^2x^3z^2]\,\binom{8}{1}\cdot 2(y-(x + 2z))^7\\=2\binom{8}{1} \binom{7}{2}[x^3z^2](-(x + 2z))^5=-2\binom{8}{1} \binom{7}{2}\binom{5}{2}2^2=-13440.$$