Number of ways to seat students at a round table subject to certain conditions. In an Olympic contest, there are $n$ teams.  Each team is composed of $k$ students attending different subjects.  How many ways are there to seat all the students at a round table such that $k$ students in a team sit together and there are no two students who attend the same subject seat next to one another?
My attempt:
Let $S_n$ denote the total way to seat all the student in $n$ teams with $k$ students on each team in a way that satisfies the problem.
Then I find that $\forall n \geq 2$ $S_{n+1}=\alpha.nS_n$ with $\alpha = 2(k-1)!-(k-2)!$
But there is problem for me to find $S_2$ because it may be non-relative to $S_1$. Help me!
 A: I came across a problem like this in a contest years ago. They have the solutions so I will give you the link. Look at question 10:
http://www.cemc.uwaterloo.ca/contests/past_contests/2009/2009EuclidContest.pdf (question paper)
http://www.cemc.uwaterloo.ca/contests/past_contests/2009/2009EuclidSolution.pdf (solution)
A formula was not needed to by derived to answer the question but somewhere in the solution they give the explicit formula.
A: I think that your recurrence isn’t quite right. If you start with an acceptable arrangement of $n$ teams, you can insert an $(n+1)$-st team in any of the $n$ slots between adjacent teams. The members of the new team can be permuted in $k!$ ways; $(k-1)!$ of these have an unacceptable person at one end, $(k-1)!$ have an unacceptable person at the other end, and $(k-2)!$ have an unacceptable person at both ends, so 
$$S_{n+1}=n\Big(k!-2(k-1)!+(k-2)!\Big)S_n\;,$$
i.e., we should have $\alpha=k!-2(k-1)!+(k-2)!$. 
I’m assuming now that arrangements that differ only by a rotation of the table are considered the same. Then $S_1=(k-1)!$.  There are at least two ways to see that $S_2=k!\alpha$.


*

*Start with any of the $(k-1)!$ arrangements of one team. There are $k$ slots into which we can insert the second team, and the argument given above shows that within its slot it can be arranged in $\alpha$ ways, for a total of $(k-1)!k\alpha=k!\alpha$ arrangements.  

*There are $k!$ ways to seat the first team around half of the table, and by the argument given above there are $\alpha$ acceptable ways to seat the second team around the other half of the table.


Combine this with the recurrence $S_n=(n-1)\alpha S_{n-1}$ for $n\ge 3$, and you can easily get a closed expression for $S_n$ in terms of $n$ and $k$.
