Let's answer the following questions:
a) How many ways can 9 people be seated I) on a bench II) around a circular table?
b) How many ways can 5 men and 4 women be seated on a bench so that I.) no two people of the same gender sit next to each other?

II.) the men and women sit in separate groups (only 1 man and 1 woman sit next to each other)?
a) I. The number of permutations of 9 elements is $9!=362880$.

II. Here, the following 9 groups are clearly one to consider: abcdefghi, bcdefghia, cdefghiab, ..., iabcdefgh. Since these cyclic permutations can be done with every permutation group, the number of all possible different arrangements is $\frac{9!}{9}=8!=40320$. (We can also reason as follows: we fix one element and by permuting the remaining 8 elements, we get all possible arrangements around the table.)

b) I. They can only sit according to the scheme $f n f n f n f n f$. The 5 men can occupy the $f$-marked places in 5! ways, and the 4 women can sit in the $n$-marked places in 4! ways in each case. Therefore, they can arrange themselves in a total of $5!4!=120 \cdot 24=2880$ ways.

II. Here, they can occupy places according to 2 schemes: fffffnnnn or nnnnfffff. Therefore, the number of all possible different arrangements is $2 \cdot 5!4!=5760$.

Kovács László (Debrecen, Ref. g. II. o. t.)