Inclusion Exclusion Problem: $x_1 + x_2 + x_3 = 17$ subject to the restrictions that $x_i \leq 7$, $1 \leq i \leq 3$

Question

How many solutions are there to

$x_{1} + x_{2} + x_{3} = 17 $ where $x_{i} \leq 7$ for $1\leq i \leq 3$

This problem and solution comes from this youtube video: https://www.youtube.com/watch?v=Y0CYHMqomgI&t=475s

Solution

$ x_{i} \leq 7 \rightarrow $ $ \overline{c_{i}} = x_{i} < 8 \rightarrow $ $ c_{i} = x_{i} \geq 8 $

$N(\overline{c_{1}c_{2}c_{3}}) = N - (Nc_{1}+Nc_{2}+Nc_{3}) + N(c_1c_2) + N(c_1c_3) + N(c_2c_3) - N(c_1c_2c_3)$

I understand the rest of the solution but not this negation part. How did he find out $N(\overline{c_{1}c_{2}c_{3}}) $ ?

Do you know the Inclusion-Exclusion Principle for three sets? It says that $$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$$ Also, you should specify that we are seeking nonnegative integer solutions. — N. F. Taussig, May 21 '19 at 18:52
You could also try to explore other techniques, like generating functions, examples here and here — rtybase, May 21 '19 at 19:06
Ok thanks is he using the "Inclusion-Exclusion Principle" with other terms? — Daniel Andersson, May 21 '19 at 19:09
can i use the method here shown in the youtube video if i got an equation with x1 + x2 + x3 +x4 +x5? — Daniel Andersson, May 22 '19 at 13:45

score 1 · Answer 1 · answered May 21 '19 at 20:05

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{x_{1} = 0}^{7}\sum_{x_{2} = 0}^{7} \sum_{x_{3} = 0}^{7}\bracks{z^{17}}z^{x_{1} + x_{2} + x_{3}}} = \bracks{z^{17}}\pars{\sum_{x = 0}^{7}z^{x}}^{3} = \bracks{z^{17}}\pars{z^{8} - 1 \over z - 1}^{3} \\[5mm] = &\ \bracks{z^{17}}\pars{1 - z^{8}}^{3}\pars{1 - z}^{-3} = \bracks{z^{17}}\pars{1 - 3z^{8} + 3z^{16}}\pars{1 - z}^{-3} \\[5mm] = &\ \bracks{z^{17}}\pars{1 - z}^{-3} - 3\bracks{z^{9}}\pars{1 - z}^{-3} + 3\bracks{z^{1}}\pars{1 - z}^{-3} \\[5mm] = &\ {-3 \choose 17}\pars{-1}^{17} - 3{-3 \choose 9}\pars{-1}^{9} + 3{-3 \choose 1}\pars{-1}^{1} \\[5mm] = &\ -{19 \choose 17}\pars{-1}^{17} + 3{11 \choose 9}\pars{-1}^{9} - 3{3 \choose 1}\pars{-1}^{1} \\[5mm] = &\ \underbrace{19 \choose 17}_{\ds{171}}\ -\ 3\ \underbrace{11 \choose 9}_{\ds{55}} + 3\ \underbrace{3 \choose 1}_{\ds{3}}\ =\ \bbx{15} \end{align}

Thanks this might be a better general way to solve these types of problems. What is this method called? — Daniel Andersson, May 22 '19 at 13:43

score 0 · Answer 2 · answered May 21 '19 at 19:38

There are $15$ triples as the following

$$(3,7,7),(4,6,7),(4,7,6),(5,5,7),(5,6,6),(5,7,5),(6,4,7),(6,5,6),(6,6,5),(6,7,4),(7,3,7),(7,4,6),(7,5,5),(7,6,4),(7,7,3)$$

It is interesting that this number is $1+2+3+4+5$ as we count possible answers for $x_1$ from $3$ to $7$.

N. F. Taussig · Answer 3 · 2019-05-22T16:32:58.007

We use DeMorgan's Laws and the Inclusion-Exclusion Principle.

One of DeMorgan's Laws states that the complement of the union of a collection of sets is the intersection of their complements. $$\overline{\bigcup_{i \in I} A_i} = \bigcap_{i \in I} \overline{A_i}$$ where $I$ is some index set. For three sets, we get $$\overline{A_1 \cup A_2 \cup A_3} = \overline{A_1} \cap \overline{A_2} \cap \overline{A_3}$$ Hence, $$|\overline{A_1 \cup A_2 \cup A_3}| = |\overline{A_1} \cap \overline{A_2} \cap \overline{A_3}|$$ If $U$ is the universal set, then $$|\overline{A_1} \cap \overline{A_2} \cap \overline{A_3}| = |\overline{A_1 \cup A_2 \cup A_3}| = |U| - |A_1 \cup A_2 \cup A_3|$$ By the Inclusion-Exclusion Principle, $$|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|$$ Therefore, \begin{align*} |\overline{A_1} \cap \overline{A_2} \cap \overline{A_3}| & = |U| - (|A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|)\\ & = |U| - |A_1| - |A_2| - |A_3| + |A_1 \cap A_2| + |A_1 \cap A_3| + |A_2 \cap A_3| - |A_1 \cap A_2 \cap A_3| \end{align*} With that in mind, you should have written $$N(\overline{c_1}~\overline{c_2}~\overline{c_3}) = N - N(c_1) - N_(c_2) - N(c_3) + N(c_1c_2) + N(c_1c_3) + N(c_2c_3) - N(c_1c_2c_3)$$ where we make the following substitutions:

$N$ for $|U|$
$N(c_i)$ for $|A_i$, $1 \leq i \leq 3$
$N(c_ic_j)$ for $|A_i \cap A_j|$, $1 \leq i, j \leq 3$
$N(c_1c_2c_3)$ for $|A_1 \cap A_2 \cap A_3|$
$N(\overline{c_1}~\overline{c_2}~\overline{c_3})$ for $|\overline{A_1} \cap \overline{A_2} \cap \overline{A_3}|$

Inclusion Exclusion Problem: $x_1 + x_2 + x_3 = 17$ subject to the restrictions that $x_i \leq 7$, $1 \leq i \leq 3$

3 Answers3