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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.

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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.

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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 $x^{15}y^4$?

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  • $\begingroup$ Wait where did the 15 come from? Where is that being defined? $\endgroup$
    – Lakeland
    Commented Sep 6, 2017 at 13:41
  • $\begingroup$ @Lakeland $c_k$ is the coefficient $x^{k}y^{19-k}$. $k=15$ is what you asked for. $\endgroup$
    – Robert Z
    Commented Sep 6, 2017 at 14:14
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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 $y^4$.

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  • $\begingroup$ @Lakeland: Let's see ... Maybe you want to add your steps in the follow-up question. $\endgroup$ Commented Sep 6, 2017 at 15:25
  • $\begingroup$ Could you please expand upon 3, and how it's calculated to the solution? Still not quite sure I understand the logic behind it. 19!/4! = 121645100408832000/24 = 5068545850368000, then 19!/4! * 2^15 * 3^4? I don't get the same answer as you so I'm clearly doing something wrongly. $\endgroup$
    – Lakeland
    Commented Sep 9, 2017 at 7:18
  • $\begingroup$ @Lakelend: Recall $\binom{n}{k}=\frac{n!}{k!(n-k)!}$, so $\binom{19}{4}=\frac{19!}{4!15!}$. $\endgroup$ Commented Sep 9, 2017 at 7:51

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