正整数解的个数$x_1+x_2+x_3=100$

hbghlyj

Example 292 How many different triples $\left(x_{1},x_{2},x_{3}\right)$ of positive integers are there such that $x_{1}+x_{2}+x_{3}=100$ and $x_{1}\leqslant x_{2}\leqslant x_{3}$.

Proof. Let

$$S=\left\{\left(x_{1},x_{2},x_{3}\right):x_{i}\geqslant 1,x_{1}+x_{2}+x_{3}=100% \right\}.$$

Then $S_{3}$ acts naturally on $S$ and in each orbit of this action there is a unique $\left(x_{1},x_{2},x_{3}\right)$ such that $x_{1}\leqslant x_{2}\leqslant x_{3}$. So the question is equivalent to finding the number of orbits of this action. Note that $x_{1}$ can be any number from 1 to 98 , and $x_{2}$ any number from 1 to $99-x_{1}$, with $x_{3}$ then determined by the choices of $x_{1}$ and $x_{2}$. So

$$|S|=\sum_{x_{1}=1}^{98}\left(99-x_{1}\right)=99\times 98-\frac{1}{2}\times 98% \times 99=49\times 99=4851$$

We apply the orbit counting formula as below:

$g$	conjugates	$s$ fixed by $g$	$\operatorname{fix}(g)$	contribution
$e$	1	$\left(x_{1},x_{2},x_{3}\right)$	4851	4851
$(12)$	3	$\left(x_{1},x_{1},x_{3}\right)$	49	147
$(123)$	2	$\left(x_{1},x_{1},x_{1}\right)$	0	0

Hence the number of orbits (and our answer) equals

$$\frac{4851+147}{6}=833\text{. }$$

$\blacksquare$

Copy tex source
Made with LaTeXML
Made with mathpix
From 2014 Notes page 79