What is the coefficient of $ a^8b^4c^9d^9$ in $(abc+abd+acd+bcd)^{10}$? 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?
 A: 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 :
$$\frac{10!}{2!\, 6!\, 1!\, 1!}$$
A: 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  $$d^8\binom82(c^2+2cd+d^2)$$
A: 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
$$ \frac{10!}{2!6!}=\color{red}{2520}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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
$$
{10! \over 1!\,1!\,6!\,2!} = {10 \times 9 \times 8 \times 7 \over 2} =
\bbox[#ffd,15px,border:1px groove navy]{\ds{2520}}
$$
