Find the coefficient of $x^2y^2z^4$ in the expansion of $(x-2y+z)^8$.

Question

I need help solving this question​. How do I start?

Attempt at solution *The term that contains $x^2y^2z^4$ is ${8 \choose 4} (-2y)^2(z)$ ? *

Answer 1

The multinomial theorem will tell you the following (I'll state it for trinomials) $$ (x_1 + x_2 + x_3)^n = \sum_{k_1 + k_2 + k_3 = n} \binom{n}{k_1,k_2,k_3} \prod_{i=1}^3 x_i^{k_i} $$

In our case, we have $x_1 = x$, $x_2 = -2y$, $x_3 =z$. So if we are finding the coefficient of $x^2 y^2 z^4$, note that this is equivalent to $x_1^2\times \frac{x_2^2}{4} \times x_3^4$, which means we have to take $k_1 = 2,k_2 = 2, k_3 = 4$. This gives us the answer $\binom{8}{2,4,4} \times 4$ (which comes because $y^2 = \frac{x_2^2}{4}$, so we have to account for this): $$ \frac{8! \times 4}{ 4! \times 2!\times 2!} = 1680 $$ which should be correct (and is , I checked on Wolfram Alpha to make sure).

I case you are not aware of this theorem, read it up on Wikipedia. The proof is similar to that of the binomial theorem.

Answer 2

We can also iteratively apply the binomial theorem in order to find the coefficients. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$.

We obtain \begin{align*} [x^2y^2z^4](x-2y+z)^8&=[x^2y^2z^4]\sum_{j=0}^8\binom{8}{j}x^j(-2y+z)^{8-j}\tag{1}\\ &=[y^2z^4]\binom{8}{2}(-2y+z)^6\\ &=\binom{8}{2}[y^2z^4]\sum_{j=0}^6\binom{6}{j}(-2y)^jz^{6-j}\tag{2}\\ &=\binom{8}{2}\binom{6}{2}(-2)^2\\ &=1680 \end{align*}

Comment:

• In (1) we apply the binomial theorem and select the coefficient of $x^2$.

• In (2) we again apply the binomial theorem and select the coefficient of $y^2z^4$.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \bracks{x^{2}y^{2}z^{4}}\pars{x - 2y + z}^{8} & = \bracks{x^{2}y^{2}z^{4}}\sum_{a = 0}^{\infty}\sum_{b = 0}^{\infty} \sum_{c = 0}^{\infty}{8! \over a!\,b!\,c!}\bracks{a + b + c = 8}x^{a}\pars{-2y}^{b}z^{c} \\[5mm] & = {8! \over 2!\,2!\,4!}\pars{-2}^{2} = 8 \times 7 \times 6 \times 5 = \bbx{\ds{1680}} \end{align}

Answer 4

Hint: if you expand $(x-2y+z)^2$, every term has total degree $2$.

Answer 5

First we need the coefficient of $x^2$ in $$\{x+(z-2y)\}^8$$ which is $$\binom86(z-2y)^6$$

Now what is the coefficient of $z^4$ in $$(z-2y)^6?$$