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I am new to Binomial Theorem and I want to find out the coefficient of $ a^8b^4c^9d^9$ in $$(abc+abd+acd+bcd)^{10}$$ How to find that?

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    $\begingroup$ As a first step: For what integers $p,q,r,s$ do you have $(abc)^{p}(abd)^{q}(acd)^{r}(bcd)^{s} = a^8b^4c^9d^9$? $\endgroup$ – Alex Zorn Jan 5 '18 at 3:13
  • $\begingroup$ @Alex Zorn $ p={ 2,2,4} \ q={2,1,1} \ r={3,3,3} \ s={3,3,3} $ $\endgroup$ – Ravi Prakash Jan 5 '18 at 3:19
  • $\begingroup$ @Alex Zorn Can you compile the next steps in answer? I like your way $\endgroup$ – Ravi Prakash Jan 5 '18 at 3:22
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Divide by $abcd$ inside the bracket to get to a simplification

$$(abcd)^{10}(a^{-1}+b^{-1}+c^{-1}+d^{-1})^{10}$$

Thus you need to search for powers $2,6,1,1$ of $a^{-1}, b^{-1}, c^{-1}, d^{-1}$ respectively to get the original expression.

The answer is given by multinomial theorem is :

$$\frac{10!}{2!\, 6!\, 1!\, 1!}$$

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  • $\begingroup$ Hey, can you please explain the expression, you got using multinomial theorem? (Edit In the answer) $\endgroup$ – Ravi Prakash Jan 5 '18 at 3:24
  • $\begingroup$ The steps before last line are convincing. You just need to tell why multinomial is used here $\endgroup$ – Ravi Prakash Jan 5 '18 at 3:25
  • $\begingroup$ @Ravi you need to search for the term containing said powers of $a^{-1}, b^{-1} ...$ in the said expansion. This is given by multinomial theorem $\endgroup$ – SJ. Jan 5 '18 at 3:29
  • $\begingroup$ Okay that is $$ \sum_{p+q+r+s=n} \frac{n!}{p!q!r!s!} x^py^qz^ru^s$$ but where is $$ \sum_{p+q+r+s=n}$$ in your answer? $\endgroup$ – Ravi Prakash Jan 5 '18 at 3:32
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – SJ. Jan 5 '18 at 3:32
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Hint:

We can use Binomial Theorem recursively

The coefficient of $a^8$ in

$$(a(bc+bd+cd)+bcd))^{10}$$ will be $$\binom{10}2(bc+bd+cd)^8(bcd)^2$$

The coefficient of $b^4$ in $$(bc+bd+cd)^8(bcd)^2$$

= the coefficient of $b^2$ in $$(b(c+d)+cd)^8(cd)^2$$

$$=(cd)^2\binom82(cd)^{8-2}(c+d)^2=c^8d^8\binom82(c^2+2cd+d^2)$$

Clearly, the coefficient of $c^9$ in $$c^8d^8\binom82(c^2+2cd+d^2)$$

= the coefficient of $c$ in $$d^8\binom82(c^2+2cd+d^2)$$

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Let us set $abc=D, abd=C, acd=B, bcd=A$. Then $a^8 b^4 c^9 d^9 =A^2 B^6 C D$ and the problem boils down to finding the coefficient of $A^2 B^6 C D$ in $(A+B+C+D)^{10}$, i.e. to counting the anagrams of the word $AABBBBBBCD$. They are $$ \frac{10!}{2!6!}=\color{red}{2520}.$$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\pars{abc + abd + acd + bcd}^{10}}} \\[5mm] = &\ \sum_{i,j,k,\ell\ \in\ \mathbb{N}_{\,\geq 0}} {10! \over i!\,j!\,k!\,\ell!}\,\pars{abc}^{i}\pars{abd}^{j}\pars{acd}^{k} \pars{bcd}^{\ell}\,\bracks{i + j + k + \ell = 10} \\[5mm] = &\ \sum_{i,j,k,\ell\ \in\ \mathbb{N}_{\,\geq 0}} {10! \over i!\,j!\,k!\,\ell!}\, a^{i + j + k}\,b^{i + j + \ell}\,c^{i + k + \ell}\,d^{j + k + \ell} \,\bracks{i + j + k + \ell = 10} \end{align}

Then,

$$ \left\{\begin{array}{rcrcrcrcrr} \ds{i} & \ds{+} & \ds{j} & \ds{+} & \ds{k} & \ds{+} & \ds{\ell} & \ds{=} & \ds{10} & \ds{\qquad\pars{\texttt{a}}} \\[1mm] \ds{i} & \ds{+} & \ds{j} & \ds{+} & \ds{k} &&& \ds{=} & \ds{8} & \ds{\qquad\pars{\texttt{b}}} \\[1mm] \ds{i} & \ds{+} & \ds{j} &&& \ds{+} & \ds{\ell} & \ds{=} & \ds{4} & \ds{\qquad\pars{\texttt{c}}} \\[1mm] \ds{i} &&& \ds{+} & \ds{k} & \ds{+} & \ds{\ell} & \ds{=} & \ds{9} & \ds{\qquad\pars{\texttt{d}}} \\[1mm] && \ds{j} & \ds{+} & \ds{k} & \ds{+} & \ds{\ell} & \ds{=} & \ds{9} & \ds{\qquad\pars{\texttt{e}}} \end{array}\right. $$

Note that $\ds{"\pars{\texttt{a}} - \pars{\texttt{b}}" \implies \bbx{\ell = 2}}$ such that

$$ \left\{\begin{array}{rcrcrcrr} \ds{i} & \ds{+} & \ds{j} & \ds{+} & \ds{k} & \ds{=} & \ds{8} & \ds{\qquad\pars{\texttt{b}}} \\[1mm] \ds{i} & \ds{+} & \ds{j} &&& \ds{=} & \ds{2} & \ds{\qquad\pars{\texttt{c}}} \\[1mm] \ds{i} &&& \ds{+} & \ds{k} & \ds{=} & \ds{7} & \ds{\qquad\pars{\texttt{d}}} \\[1mm] && \ds{j} & \ds{+} & \ds{k} & \ds{=} & \ds{7} & \ds{\qquad\pars{\texttt{e}}} \end{array}\right. $$

Note that $\ds{"\pars{\texttt{b}} - \pars{\texttt{c}}" \implies \bbx{k = 6}}$ such that $\ds{\bbx{i = j = 1}}$ from $\ds{\pars{\texttt{d}}}$ and $\ds{\pars{\texttt{e}}}$.

The coveted coefficient is given by $$ {10! \over 1!\,1!\,6!\,2!} = {10 \times 9 \times 8 \times 7 \over 2} = \bbox[#ffd,15px,border:1px groove navy]{\ds{2520}} $$

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  • $\begingroup$ How to suffer needlessly... $\endgroup$ – Did Jan 8 '18 at 18:19
  • $\begingroup$ @Did Happy New Year 2018. $\endgroup$ – Felix Marin Jan 9 '18 at 18:40

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