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My math teacher was explaining a question in the chapter Combinatorics and I met with a hiccup in the solution. The question is given below and the solution (as given by my teacher) is also given in the form of a screenshot of the zoom meeting.

$3$ dices are rolled simultaneously once and the resulting numbers are added. Find the number of ways of getting a total of $10$.

The solution

The hiccup is that I cant understand how the last line is equal to the one before it.

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    $\begingroup$ Please don't use pictures. $\endgroup$ Nov 17 '20 at 15:45
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    $\begingroup$ They are certainly not equal. I think your teacher simply omitted the terms with higher powers ($x^{12}, x^{18}$) since they do not contribute to the coefficient of $x^7$. $\endgroup$
    – player3236
    Nov 17 '20 at 15:51
  • $\begingroup$ Thanks a lot.... And I am new to the community so I cant use MathJax that's why I had to use picture links, will keep that in mind. $\endgroup$ Nov 17 '20 at 16:21
  • $\begingroup$ @SupragyaMishra: Here is a basic MathJax tutorial. $\endgroup$ Nov 17 '20 at 18:55
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    $\begingroup$ Thanks @BrianM.Scott $\endgroup$ Nov 19 '20 at 11:46
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$


\begin{align} &\bbox[5px,#ffd]{\sum_{d_{1} = 1}^{6}\sum_{d_{2} = 1}^{6}\sum_{d_{3} = 1}^{6} \bracks{z^{10}}z^{d_{1}\ +\ d_{2}\ +\ d_{3}}} = \bracks{z^{10}}\pars{\sum_{d_{1} = 1}^{6}z^{d}}^{3} = \bracks{z^{10}}\pars{z\,{z^{6} - 1 \over z - 1}}^{3} \\[5mm] = &\ \bracks{z^{7}}\pars{1 - z^{6}}^{3}\pars{1 - z}^{-3} = \bracks{z^{7}}\pars{1 - 3z^{6}}\pars{1 - z}^{-3} \\[5mm] = &\ \bracks{z^{7}}\pars{1 - z}^{-3} - 3\bracks{z^{1}}\pars{1 - z}^{-3} = {-3 \choose 7}\pars{-1}^{7} - 3{-3 \choose 1}\pars{-1}^{1} \\[5mm] = &\ {9 \choose 7} - 3{3 \choose 1} = 36 - 3 \times 3 = \bbx{27} \\ & \end{align}
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