How many solutions are there to the equation $$ x_{1} + x_{2} + x_{3} + x_{4} = 28\ {\large ?}\qquad\mbox{if}\qquad x_{1} \geq 3\,,\ x_{2} \geq 3\,,\ x_{3} \geq 5\ \mbox{and}\ x_{4} \geq 5. $$ $x_{1}, x_{2}, x_{3}, x_{4}$ positive odd numbers.
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Have you tried transforming the $x_i$ to $y_i$ so that you get something like $y_1+y_2+y_3+y_4=n$ for $y_i$ non-negative integers? – Arthur Dec 14 '14 at 09:59
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$\displaystyle 84$. – Felix Marin Dec 15 '14 at 18:22
2 Answers
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HINT:
Let $y_1=x_1-3,y_2=x_2-3,y_3=x_3-5,y_4=x_4-5$
$\implies\sum_{i=1}^4 y_i=12$
As $y_i$s are even, let $y_i=2z_i$ where integer $z_i\ge0$
$\implies\sum_{i=1}^4(2z_i)=12\iff\sum_{i=1}^4 z_i=6$
lab bhattacharjee
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once question again, How many solutions are there to the equation a+b+c=15, abc is integer, 100<abc<900 – nalmar fendi Dec 14 '14 at 15:50
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@RamlanEffendi, Why don't you put this as an independent question to invite even greater audience. – lab bhattacharjee Dec 14 '14 at 15:53
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i cannot type this as an independent question. maybe something wrong. i will try again – nalmar fendi Dec 14 '14 at 16:02
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every i put as an independent question, there is an error. like that : "Oops! Your question couldn't be submitted because: • This question body does not meet our quality standards. Please make sure that it completely describes your problem - including what you have already tried - and is written using proper grammar. – nalmar fendi Dec 14 '14 at 16:37
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@FelixMarin It is well-known and easy to see that it equals the number of ways to select $3$ (separating) cells in a strip of length $6+3$, that is ${9 \choose 3}$, but you have already found a way to this number. :-) – Alex Ravsky Aug 05 '17 at 05:52
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \sum_{x_{1}\ =\ 3 \atop x_{1}\ {\rm odd}}^{\infty} \sum_{x_{2}\ =\ 3 \atop x_{2}\ {\rm odd}}^{\infty} \sum_{x_{3}\ =\ 5 \atop x_{3}\ {\rm odd}}^{\infty} \sum_{x_{4}\ =\ 5 \atop x_{4}\ {\rm odd}}^{\infty} \delta_{x_{1} + x_{2} + x_{3} + x_{4},28}} \\[5mm]&=\sum_{a\ =\ 0}^{\infty}\sum_{b\ =\ 0}^{\infty}\sum_{c\ =\ 0}^{\infty} \sum_{d\ =\ 0}^{\infty} \delta_{\pars{2a + 3} + \pars{2b + 3} + \pars{2c + 5} + \pars{2d + 5},28} =\sum_{a\ =\ 0}^{\infty}\sum_{b\ =\ 0}^{\infty}\sum_{c\ =\ 0}^{\infty} \sum_{d\ =\ 0}^{\infty}\delta_{a + b + c + d,6} \\[5mm]&=\sum_{a\ =\ 0}^{\infty}\sum_{b\ =\ 0}^{\infty}\sum_{c\ =\ 0}^{\infty} \sum_{d\ =\ 0}^{\infty}\oint_{\verts{z}\ =\ 1^{-}}{1 \over z^{-a - b - c - d + 7}} \,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ 1^{-}}{1 \over z^{7}}\pars{\sum_{a\ =\ 0}^{\infty}z^{a}}^{4} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1^{-}}{1 \over z^{7}}\,\pars{1 \over 1 - z}^{4} \,{\dd z \over 2\pi\ic} =\sum_{k\ =\ 0}^{\infty}{-4 \choose k}\pars{-1}^{k}\ \overbrace{% \oint_{\verts{z}\ =\ 1^{-}}{1 \over z^{7 - k}}\,{\dd z \over 2\pi\ic}} ^{\dsc{\delta_{k,6}}}\ =\ {-4 \choose 6}\pars{-1}^{6} \\[5mm]&={-\bracks{-4} + 6 - 1 \choose 6}\pars{-1}^{6}={9 \choose 6} ={9 \times 8 \times 7 \over 3 \times 2}=\color{#66f}{\Large 84} \end{align}
Felix Marin
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This is great! I think you may provide a link to your answer to this thread. :-) – Alex Ravsky Aug 05 '17 at 05:54