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.
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Is this not including triangles with 2 or more equal sides? Or should those "less than"s be "less than or equal to"? – Mike May 09 '17 at 22:32
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It's less than as I wrote $a<b<c$.@Mike – Iti Shree May 09 '17 at 22:33
2 Answers
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$c$ 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 $7$
Ross Millikan
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So if I cross across such question shall I hand count? Thanks by the way. – Iti Shree May 09 '17 at 22:29
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$21$ is a small number and the strict less than signs reduce the number of possibilities. One has to judge the best approach and this seemed reasonable. Even with typing this took less than a minute. – Ross Millikan May 09 '17 at 22:52
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
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.
Felix Marin
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