Aside from $1!\cdot n!=n!$ and $(n!-1)!\cdot n! = (n!)!$, the only nontrivial product of factorials known is $6!\cdot 7!=10!$.
One might naturally associate these numbers with the permutations on $6, 7,$ and $10$ objects, respectively, and hope that this result has some kind of connection to a sporadic relation between such permutations - numerical "coincidences" often have deep math behind them, like how $1^2+2^2+\ldots+24^2=70^2$ can be viewed as an ingredient that makes the Leech lattice work.
The most natural thing to hope for would be a product structure on the groups $S_6$ and $S_7$ mapping to $S_{10}$, but as this MathOverflow thread shows, one cannot find disjoint copies of $S_6$ and $S_7$ living in $S_{10}$, so a product structure seems unlikely.
However, I'm holding out hope that some weaker kind of bijection can be found in a "natural" way. Obviously one can exhibit a bijection. For instance, identify the relative ordering of $1,2,\ldots 7$ in a permutation of size $10$, and then biject $_{10}P_{3}=720$ with $S_6$ in some way. But I'd like to know if there is a way to define such a bijection which arises naturally from the permutation structures on these sets, and makes it clear why the construction does not extend to other orders.
I tried doing something with orderings on polar axes of the dodecahedron ($10!$) and orderings on polar axes of the icosahedron ($6!$), in the hopes that the sporadic structure and symmetry of these Platonic solids would allow for interesting constructions that don't generalize, but ran into issues with the dodecahedron (sequences of dodecahedral axes aren't particularly nice objects) and the question of how to extract a permutation of length $7$.
I'm curious if someone can either devise a natural bijection between these sets or link to previous work on this question.
from Hot Weekly Questions - Mathematics Stack Exchange
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