What is the coefficient of $ a^8b^4c^9d^9$ in $(abc+abd+acd+bcd)^{10}$?

Question

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?

Answer 1

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 :

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

Answer 2

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 $$d^8\binom82(c^2+2cd+d^2)$$

Answer 3

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 $$ \frac{10!}{2!6!}=\color{red}{2520}.$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $$ {10! \over 1!\,1!\,6!\,2!} = {10 \times 9 \times 8 \times 7 \over 2} = \bbox[#ffd,15px,border:1px groove navy]{\ds{2520}} $$