Binomial Expansion coefficients Find the coefficient of $x^3 y^4$ in the expansion of $(2x-4y)^7$. I would also like an explanation for how the final answer was obtained.
 A: The Binomial Theorem states 
$$(a+b)^n= \sum_{r=0}^{n}\binom{n}{ r} (a)^{n-r}(b)^r$$
and so
$$(2x-4y)^7=\sum_{r=0}^{7}\binom{7}{ r} (2x)^{7-r}(-4y)^r.$$
So, we simply need to consider the $5$th term  (i.e., when $r=4$):
$$\binom{7}{ 4} (2x)^{7-4}(-4y)^4=\binom{7}{4}2^34^4x^3y^4.$$
I’ll leave it to you to evaluate what this equals.
In general, whenever you are asked to find a particular coefficient in the Binomial expansion, try to reword the question into one which asks “What value of $r$ will give me the desired term?” and then evaluate $$\binom{n}{r}(a)^{n-r}(b)^{r}$$ for that particular $r$, as I did above.
A: It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$.

We obtain
  \begin{align*}
\color{blue}{[x^3y^4]}&\color{blue}{(2x-4y)^7}\\
&=[x^3y^4]\sum_{k=0}^7\binom{7}{k}(2x)^k(-4y)^{7-k}\tag{1}\\
&=[y^4]\binom{7}{3}2^3(-4y)^4\tag{2}\\
&=\binom{7}{3}2^3(-4)^4\tag{3}\\
&\,\,\color{blue}{=71\,680}
\end{align*}

Comment:


*

*In (1) we apply the binomial theorem.

*In (2) we select the coefficient of $x^3$.

*In (3) we select the coefficient of $x^4$.
