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What is the coefficient of $x^2y^2z^3$ in $(x + 2 y + z)^7 $?

This is the question at a test and the correct answer is given as 840.

Isn't it $7!/(2!2!3!)$ ?

Alexandru Cimpanu
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2 Answers2

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The coefficient of $x^2y^2z^3$ in $(x+y+z)^7$ is indeed $\frac{7!}{2!\cdot2!\cdot3!}=210$

however here we have $2y$ instead of $y$. So instead of $y^2$ we have $4y^2$. Multiplying your result by $4$ gives the desired result.

Asinomás
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  • what would be the general solution? I will have to multiply the result by what? 2 to the power of y in our case? – Alexandru Cimpanu Jul 07 '14 at 14:26
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    the multinomial theorem allows you to compute the external coefficient, you then need to calculate what the internal coefficient of each of the powers will be and multiply the external coefficient with all the internal coefficients, this will give you the result. Notice in this example the external coefficient was 210, and all the external coefficients were 1 except for the coefficient for the $y^2$ term, which was $4$. – Asinomás Jul 07 '14 at 14:47
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \pars{x + 2y + z}^{7}& &=\sum_{{\vphantom{\LARGE A}a,b,c=0}\atop {\vphantom{\huge A}a + b + c = 7}}^{\infty} {7! \over a!\,b!\,c!}\,x^{a}\pars{2y}^{b}z^{c} =\sum_{{\vphantom{\LARGE A}a,b,c=0}\atop {\vphantom{\huge A}a + b + c = 7}}^{\infty} {7! \over a!\,b!\,c!}\,2^{b}\pars{x^{a}y^{b}z^{c}} \end{align}

$$ {7! \over 2!\,2!\,3!}\,2^2 = 7\cdot 6\cdot 5\cdot 4 = \color{#66f}{\large 840} $$

Felix Marin
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