The number of triplets (a,b,c) of integers such that $aI have tried getting possible combinations where $a<b<c$ but it's taking too long and since this is a objective question for competitive exam (source, problem 43), I want to know any easy method to solve it.
 A: $c$ has to be at least $8$ and can not be more than $10$, so a hand count is not very long.  I find $(6,7,8),(5,7,9), (4,8,9), (5,6,10), (4,7,10),(3,8,10), (2,9,10) $ for $7$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

By definition, the answer is given by

\begin{align}
&\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}\sum_{c = 1}^{\infty}
\bracks{a < b < c}\bracks{b - a \leq c \leq a + b}\bracks{z^{21}}z^{a + b + c}
\\[5mm] = &\
\bracks{z^{21}}\sum_{a = 1}^{\infty}z^{a}
\sum_{b = 1}^{\infty}\bracks{b \geq a + 1}z^{b}\sum_{c = 1}^{a + b}
\bracks{c \geq b + 1}\bracks{c \geq b - a}z^{c}
\\[5mm] = &\
\bracks{z^{21}}\sum_{a = 1}^{\infty}z^{a}
\sum_{b = a + 1}^{\infty}z^{b}\sum_{c = b + 1}^{a + b}z^{c} =
\bracks{z^{21}}\sum_{a = 1}^{\infty}z^{a}
\sum_{b = a + 1}^{\infty}z^{b}\pars{z^{b + 1}\,{z^{a} - 1 \over z - 1}}
\\[5mm] = &\
\bracks{z^{21}}{z\ \over 1 - z}\sum_{a = 1}^{\infty}\pars{z^{a} - z^{2a}}
\sum_{b = a + 1}^{\infty}\pars{z^{2}}^{b} =
\bracks{z^{20}}{1 \over 1 - z}\sum_{a = 1}^{\infty}\pars{z^{a} - z^{2a}}\,
{\pars{z^{2}}^{a + 1} \over 1 - z^{2}}
\\[5mm] = &\
\bracks{z^{20}}{z^{2} \over \pars{1 - z}\pars{1 - z^{2}}}
\sum_{a = 1}^{\infty}\bracks{\pars{z^{3}}^{a} - \pars{z^{4}}^{a}}
\\[5mm] = &\
\bracks{z^{18}}{1 \over \pars{1 - z}\pars{1 - z^{2}}}
\pars{{z^{3} \over 1 - z^{3}} - {z^{4} \over 1 - z^{4}}}
\\[5mm] = &\
\underbrace{%
\bracks{z^{15}}{1 \over \pars{1 - z}\pars{1 - z^{2}}\pars{1 - z^{3}}}}
_{\ds{=\ 27}}\ -\
\underbrace{%
\bracks{z^{14}}{1 \over \pars{1 - z}\pars{1 - z^{2}}\pars{1 - z^{4}}}}
_{\ds{=\ 20}}\ =\ \bbx{\large 7}
\end{align}

Why ?:

\begin{align}
&\bracks{z^{15}}{1 \over \pars{1 - z}\pars{1 - z^{2}}\pars{1 - z^{3}}} =
\bracks{z^{15}}
\sum_{a = 0}^{\infty}\sum_{b = 0}^{\infty}\sum_{c = 0}^{\infty}z^{a + 2b + 3c}
\\[5mm] = &\
\sum_{c = 0}^{\infty}\sum_{b = 0}^{\infty}\sum_{a = 0}^{\infty}
\bracks{a + 2b + 3c = 15} =
\sum_{c = 0}^{\infty}\sum_{b = 0}^{\infty}\sum_{a = 0}^{\infty}
\bracks{a = 15 - 2b - 3c}
\\[5mm] = &\
\sum_{c = 0}^{\infty}\sum_{b = 0}^{\infty}\bracks{15 - 2b - 3c \geq 0} =
\sum_{c = 0}^{\infty}\sum_{b = 0}^{\infty}\bracks{b \leq {15 - 3c \over 2}}
\\[5mm] = &\
\sum_{c = 0}^{\infty}\pars{\left\lfloor\,{15 - 3c \over 2}\,\right\rfloor + 1}
\bracks{{15 - 3c \over 2} \geq 0} =
\sum_{c = 0}^{5}\pars{\left\lfloor\,{15 - 3c \over 2}\,\right\rfloor + 1}
\\[5mm] = &\
8 + 7 + 5 + 4 + 2 + 1 = \bbx{\large 27}
\end{align}

The other one can be evaluated in a similar fashion.

