The number of triplets (a,b,c) of integers such that $a<b<c$ are sides of triangle with perimeter 21 is?

Question

I 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.

Answer 1

$c$ 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 $7$

Answer 2

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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.