I have the following exercise: "Find number of ways to sit n people in a long table with n seats":

a. Everyone is seated on one-side of table: This case it's n! (ways).

b. We assume seating left to right is the same as seating right to left (seating A, B, C, ... from left to right is the same as from right to left, we only care about who each person's neighbor is): This case it's n!/2 since we exclude the overlap cases in part a.

c. n is even and half of the people are seated on each side of table: This case I believe it is 2 * (n/2)! since there are n/2 seats and n/2 people in each side.

d. We seat people on both sides as in (c) and all we care about is who a person’s neighbors are on each side, as in (b).

e. We are dealing with a seating as in (d), but now we also care about who is sitting opposite a person as well as who a persons neighbors on each side are.

I would really appreciate if anyone could verify if my answers for part a, b, c are correct as well as give me hints for part d and e.

Thank you.

  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Feb 21 at 10:52

Your answers for (a) and (b) are alright.

Proceed in (c) as follows:

Since there are $n$ individuals, the number of ways to select the group of size $\frac{n}{2}$ to sit on one side of the table will be given by $n\choose {n/2}$ (the remaining $\frac{n}{2}$ would sit on the other side). Now once we are done splitting into two groups, we can use the logic behind (a) to obtain the answer as

$${n\choose{n/2}}\times (n/2)!\times(n/2)!$$

$$= n!$$

Proceed in (d) as follows:

The logic from (c) is to be used again. In addition, you have to now think of the cases in which neighbors will be preserved. The list of such cases for a specific arrangement has been provided below.

$\bullet$ The original arrangement

$\bullet$ Arrangement with seat reversal on one side

$\bullet$ Arrangement with seat reversal on other side

$\bullet$ Arrangement with seat reversals on both sides

It is also certain that their respective counterparts obtained by switching the sides of the table also preserve neighbors. Thus a single arrangement is being counted $8$ times.
The answer can now be obtained as

$$\bigg({n\choose{n/2}}\times (n/2)!\times (n/2)!\bigg)\times \frac{1}{8}$$


Proceed in (e) as follows:

Once again, listing out the cases where both neighbors and opposites are preserved.

$\bullet$ The original arrangement

$\bullet$ Arrangement with seat reversals on both sides

Now, switching sides for the $2$ arrangements above would also preserve neighbors and opposites. Therefore any given arrangement appears a total of $4$ times. The answer can now be obtained as $$\bigg({n\choose{n/2}}\times (n/2)!\times(n/2)!\bigg)\times \frac{1}{4}$$


  • 1
    $\begingroup$ For (e), interchanging the two sides of the table would also preserve opposites and neighbors, so I think it should be $n!/4$. Of course, I have been wrong before! $\endgroup$ – Mike Earnest Feb 20 at 23:24
  • $\begingroup$ @MikeEarnest I had not considered that case. Hope the edited answer is satisfactory. $\endgroup$ – s0ulr3aper07 Feb 21 at 5:28

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