The number of distinct permutations of $n$ objects (say $1$, $2$, . . . , $n$) around a circular table is $(n-1)!$
I was reviewing old stuff, and tried to prove this again. I need help : Is my understanding correct?
Say $S$ is the set of all distinct permutations of $n$ objects around a circular table. We are interested about the cardinality of $S$, i.e $|S|$.
Say we're viewing these permutations from the top : Each of the permutations in $S$ can be rotated in a way so that $n$ appears at the top of the table.
Now the rest $(n-1)$ objects can permute linearly to give us all the elements of $S$.
$\boxed{\therefore |S|=(n-1)!}$
