Find the coefficient of $x^{15}y^4$ in the expansion of $(2x – 3y)^{19}$ Not even sure I understand this question or know where to start, I understand the basics of finding coefficients but don't get how it's calculated "in the expansion of" another number. Looking for some sort of explanation as to how to solve this. I understand solving binomial coefficients but I'm not entirely sure how they related in this type of question.
 A: Note that by the binomial theorem
$$\begin{align}(2x – 3y)^{19}=(2x + (-3y))^{19}&=\sum_{k=0}^{19}\binom{19}{ k}(2x)^k(– 3y)^{19-k}\\&=\sum_{k=0}^{19}\left[\binom{19}k(2)^k(-3)^{19-k}\right]x^ky^{19-k}\end{align}$$
Therefore the expansion of the $19$th-power of the binomial $2x – 3y$ is a sum of terms of the form $c_kx^ky^{19-k}$.
Finding the coefficient of $x^ky^{19-k}$ means that we have to find the coefficient $c_k$. 
So what it is the coefficient $c_{15}$ of $x^{15}y^4$?
A: Do you know the binomial theorem? What does it say about $(2x-3y)^{19}$? That expanded form is what they mean by "the expansion". Once you have that expansion, somewhere in there there is a term with $x^{15}y^4$, and that term has a coefficient. You're being asked what that coefficient is.
A: In order to obtain the coefficient it is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.

We obtain
  \begin{align*}
\color{blue}{[x^{15}y^4]}\color{blue}{(2x-3y)^{19}}
&=[x^{15}y^4]\sum_{j=0}^{19}\binom{19}{j}(2x)^j(-3y)^{19-j}\tag{1}\\
&=[y^4]\binom{19}{15}2^{15}(-3y)^{4}\tag{2}\\
&=\binom{19}{4}2^{15}\cdot 3^4\tag{3}\\
&\color{blue}{=102\,877\,110\,208}
\end{align*}

Comment:


*

*In (1) we apply the binomial theorem.

*In (2) we select the coefficient of $x^{15}$.

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