One die is rolled three times One die is rolled three times. What is the probability that you get a strictly increasing sequence of rolls (meaning roll 1 < roll 2 < roll 3)? 

The total number of possible outcomes is 216. I was shown the answer to this question, but not the steps taken to get the answer. I would like to know how visually, if that makes sense. 
If it says increasing, it means I can't have, for example: (1 < 4 < 6)?
Or does it have to be in order, like: (1 < 2 < 3), (2 < 3 < 4), (4 < 5 < 6)?
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\sum_{r_{1} = 1}^{6}{1 \over 6}\sum_{r_{2} = 1}^{6}{1 \over 6}
\sum_{r_{3} = 1}^{6}{1 \over 6}\bracks{r_{1} < r_{2}}\bracks{r_{2} < r_{3}} =
{1 \over 216}\sum_{r_{1} = 1}^{6}\sum_{r_{2} = 1}^{6}
\bracks{r_{1} < r_{2}}\bracks{r_{2} < 6}\sum_{r_{3} = r_{2} + 1}^{6}1
\\[5mm] = &\
{1 \over 216}\sum_{r_{1} = 1}^{6}\sum_{r_{2} = 1}^{6}
\bracks{r_{1} < r_{2} < 6}\pars{6 - r_{2}} =
{1 \over 216}\sum_{r_{1} = 1}^{6}\bracks{r_{1} < 5}
\sum_{r_{2} = r_{1} + 1}^{5}\pars{6 - r_{2}}
\\[5mm] = &\
{1 \over 216}\sum_{r_{1} = 1}^{4}
{\pars{r_{1} - 5}\pars{r_{1} - 6} \over 2} = \bbx{5 \over 54}
\end{align}
