1

\begin{align} &\mbox{Let}\quad x_{1} + x_{2} + x_{3} + x_{4} = 20\quad \mbox{which satisfy:}\quad \left\{\begin{array}{rcccl} {\displaystyle 1} & {\displaystyle \leq} & {\displaystyle x_{1}} & {\displaystyle \leq} & {\displaystyle 6} \\ {\displaystyle 0} & {\displaystyle \leq} & {\displaystyle x_{2}} & {\displaystyle \leq} & {\displaystyle 5} \\ {\displaystyle 4} & {\displaystyle \leq} & {\displaystyle x_{3}} & {\displaystyle \leq} & {\displaystyle 9} \\ {\displaystyle 2} & {\displaystyle \leq} & {\displaystyle x_{4}} & {\displaystyle \leq} & {\displaystyle 7} \end{array}\right. \\ &\mbox{Determine the number of integral solutions.} \\ &\mbox{} \end{align}

I figured out that I could change each equation to some $y_{i}$ such that $0 \leq y_{i} \leq 6$.
So I would have $y_{1} = x_{1}, y_{2} = x_{2} + 1, y_{3} = x_{3} - 3$ and $y_{4} = x_{4} - 1$.
This then gives me an equation where $y_{1} + \cdots + y_{4} = 17$.

From here, I do not know where to go. How do I figure out the number of integer solutions ?.

Felix Marin
  • 89,464
sunnyjim2479
  • 115

2 Answers2

1

The first step is to transform the bounds on your variables. The equivalent problem is to find the number of integer solutions for:

$z_1+z_2+z_3+z_4=17 \\ 1\leq z_1,z_2,z_3,z_4\leq 6$

Now let $y_i=7-z_i$, then an auxiliary problem is

$y_1+y_2+y_3+y_4=11 \\ 1\leq y_1,y_2,y_3,y_4\leq 6$

We can now start a stars-and-bars approach. We're going to be separating eleven stars using three bars. That's $_{10}C_3$. Now, we're going to subtract off the ways we can go outside of our bounds with any of the variables.

Suppose one of our variables is bigger than $6$. WLOG, let $y_1>6$. There are two valid cases, one when $y_1=7$ and one when $y_1=8$. We can't have $y_1$ be bigger than that, because $3$ units must be left over for the other variables.

When $y_1=7$, we count arrangements by applying the stars-and-bars approach to the remaining four units. That's $_3C_2$. When $y_1=8$, there's only going to be one valid arrangement for the other variables, $y_2=y_3=y_4=1$. That's a total of $_3C_2+1=3+1=4$ infringing arrangements when $y_1>6$. Because our choice of infringing variable was arbitrary, we need to account for the other three variables, too. That's a total of $4\times4=16$ infringing arrangements. No two variables can be over six at the same time, so there's no double-counting.

The total number of valid solutions to this problem would then be $_{10}C_3-16=120-16=104$.

Benjamin Roycraft
  • 464
1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{% \sum_{x_{1}\ =\ 1}^{6}\sum_{x_{2}\ =\ 0}^{5}\sum_{x_{3}\ =\ 4}^{9} \sum_{x_{4}\ =\ 2}^{7}\delta_{x_{1} + x_{2} + x_{3} + x_{4},\,20}}} = \sum_{x_{1}\ =\ 0}^{5}\sum_{x_{2}\ =\ 0}^{5}\sum_{x_{3}\ =\ 0}^{5} \sum_{x_{4}\ =\ 0}^{5} \delta_{\pars{x_{1} + 1} + x_{2} + \pars{x_{3} + 4}+ \pars{x_{4} + 2},20} \\[5mm] = &\ \sum_{x_{1}\ =\ 0}^{5}\sum_{x_{2}\ =\ 0}^{5}\sum_{x_{3}\ =\ 0}^{5} \sum_{x_{4}\ =\ 0}^{5}\delta_{x_{1} + x_{2} + x_{3} + x_{4},13} = \sum_{x_{1}\ =\ 0}^{5}\sum_{x_{2}\ =\ 0}^{5}\sum_{x_{3}\ =\ 0}^{5} \sum_{x_{4}\ =\ 0}^{5}\bracks{z^{13}}z^{x_{1} + x_{2} + x_{3} + x_{4}} = \bracks{z^{13}}\pars{\sum_{x = 0}^{5}z^{x}}^{4} \\[5mm] = &\ \bracks{z^{13}}\pars{z^{6} - 1 \over z - 1}^{4} = \bracks{z^{13}}\pars{1 - z^{6}}^{4}\pars{1 - z}^{-4} = \bracks{z^{13}}\sum_{k = 0}^{4}{4 \choose k}\pars{-z^{6}}^{k} \sum_{n = 0}^{\infty}{-4 \choose n}\pars{-z}^{n} \\[5mm] = &\ \bracks{z^{13}}\sum_{k = 0}^{4}\sum_{n = 0}^{\infty}{4 \choose k} {-4 \choose n}\pars{-1}^{k + n}\,z^{6k + n} = \sum_{k = 0}^{4}\sum_{n = 0}^{\infty}{4 \choose k} {n + 3 \choose n}\pars{-1}^{k}\,\,\delta_{6k + n,13} \\[5mm] = &\ \sum_{k = 0}^{4}\sum_{n = 0}^{\infty}{4 \choose k} {n + 3 \choose 3}\pars{-1}^{k}\,\,\delta_{n,13 - 6k} = \sum_{k = 0}^{4}{4 \choose k} {16 - 6k \choose 3}\pars{-1}^{k}\bracks{13 - 6k \geq 0} \\[5mm] = &\ \sum_{k = 0}^{4}{4 \choose k} {16 - 6k \choose 3}\pars{-1}^{k}\bracks{k \leq {13 \over 6}} = \sum_{k = 0}^{2}{4 \choose k}{16 - 6k \choose 3}\pars{-1}^{k} \\[5mm] = &\ \underbrace{{4 \choose 0}{16 \choose 3}}_{\ds{560}}\ -\ \underbrace{{4 \choose 1}{10 \choose 3}}_{\ds{480}}\ +\ \underbrace{{4 \choose 2}{4 \choose 3}}_{\ds{24}}\ =\ \bbox[10px,border:2px groove navy]{\ds{\large 104}} \end{align}

Felix Marin
  • 89,464