I am asked to find the number of distinct ways to express 1,000 as a product of three integers using Burnside's Lemma. By distinct ways, I mean that the order of the factors is not important, e.g., $2 \cdot 4 \cdot 125$ is not distinct from $125 \cdot 4 \cdot 2$. I am aware that there are several similar questions to this, but none that I see use Burnside's Lemma:
$$|X / G| = \frac{1}{|G|} \sum_{g \in G} |X^{g}|.$$
Here $|X / G|$ is the number of $G$ orbits and $X^{a}$ is the number of points fixed by a permutation $a$ in $G$.
I understand the lemma and its proof, but I am not sure what group and group action is relevant to the number 1,000 for this problem. I'm fairly certain that I have to consider the prime factorization: $2^{3}5^{3}$.
Any help on making sense of this question is greatly appreciated.
Best Answer
I have been able to come to an answer with the help of one of the comments. We consider the set $X$ to be the set of all (not necessarily unique) products of $3$ integers that equal $1000$. We express these products as $3$-tuples. The group $S_{3}$ acts on this set by permuting the elements in each $3$-tuple.
We first observe that the cardinality of $X$ is $100$ after considering the method from my comment to my own post. We see that the identity of $S_{3}$ fixes all $100$ elements of $X$. We see that each of the three transpositions individually fix four elements of $X$ giving us $12$ more fixed points. We then observe that the two $3$-cycles only fix the tuple $(10,10,10)$ giving us two more fixed points.
$$ \dfrac{100 + 4 + 4 + 4 + 1 + 1}{3!} = 19 $$
This number is the same as my "brute force" calculation, so I am reasonably confident in my answer.