# People seated around two tables with numbered chairs

In how many ways $3k$ people can be seated on numbered chairs around two round tables? First table have $2k$ chairs, second table have $k$ chairs. In how many ways they can be seated if two fixed persons have to sit abreast? Assume that $k \ge 2$

I tried to solve it but it turned out that my solution is wrong.

HINT:

• How many ways are there to choose $2k$ of the $3k$ people to sit at the first table?
• Now imagine that the seats at the first table are numbered from $1$ to $2k$. In how many different orders can the $2k$ people chosen for the first table sit in those seats? Note that you can specify a seating by listing the $2k$ people in the order in which they appear in seats $1$ through $2k$.
• In how many orders can the remaining $k$ people sit in the $k$ seats at the second table?
• Finally, what do you have to do with these three numbers in order to get the final answer?

Added (and corrected): To handle the second question, split it into two cases depending on whether the specified persons sit at the first table or the second. If they sit at the first, there are $2k$ pairs of adjacent seats that they can choose, and they can sit in them in either order, so there are $4k$ ways for them to seat themselves. The remaining $3k-2$ people can then seat themselves in any order in the remaining $3k-2$ seats. If they sit at the second table, there are $k$ pairs of adjacent seats that they can choose, and they can sit in them in either order, so there are $2k$ ways for them to seat themselves, and the remaining $3k-2$ people can again seat themselves in any order in the remaining $3k-2$ seats.

Note, though, that this analysis fails for $k=1$ and $k=2$. For $k=1$ the specified pair must sit at the first table, so there are only $2$ possible seatings. For $k=2$ the first part of the analysis is fine: if they sit at the first table, there are $8$ ways for them to seat, and the remaining $4$ people can sit in any of the remaining $4$ seats. If they sit at the second table, however, there are only $2$ possible arrangements for them, not $4$.

• Is the answer $(3k)!$ ? – user4201961 Oct 14 '15 at 20:14
• @user4201961: Yes, it is. By the way, I just realized that there’s a shorter way to arrive at that result: each way of seating them is uniquely obtained by lining them up in any of the $(3k)!$ possible orders, then having the first $2k$ take seats $1$ through $2k$ at the first table in that order, and finally having the last $k$ take seats $1$ through $k$ at the second table in that order. – Brian M. Scott Oct 14 '15 at 20:18
• Thanks. What about the second part? "In how many ways they can be seated if two fixed persons have to sit abreast?" – user4201961 Oct 14 '15 at 20:18
• @user4201961: Sorry, I missed that part. Let me add something to my answer. Okay, done; see if that helps. – Brian M. Scott Oct 14 '15 at 20:20
• @user4201961: Yes, for $k\ge 3$: the analysis breaks down for $k=1$ and $k=2$, and I’ll add a note to that effect. – Brian M. Scott Oct 15 '15 at 20:35