Could anybody answer this?

Seven guests turned up for dinner at a round table with 10 seats. If no empty seats are removed and no empty seats are conseccutive, then the number of ways of seating the guests is:





  • 1
    $\begingroup$ This depends on what kind of different seatings the group considers "the same". Do they, for instance, care who sits closest to the door / furthest to the north, or is that entirely irrelevant? Do they care about right and left, or only which two people (or empty seat) each of them sit between? $\endgroup$
    – Arthur
    Feb 17 at 12:53

Well, there are $7! = 5040$ distinct orderings of the seven guests (assuming there are no identical twins/ triplets etc. amongst them !). And there are $10 \times 9 \times 8 = 720$ ways of choosing three empty seats ...

Oh, wait, empty seats must not be consecutive - so there are $10$ ways to choose the first empty seat, then $7$ choices for the second empty seat. And the number of choices for the third empty seat depends on how close the first and second are. So there are

$10 \times (2 \times 5 + 5 \times 4) = 300$

ways to place the empty seats. But the empty seats are identical to one another, so we must divide this total by $3!$ to get $50$ distinct placings for the empty seats.

So our total number of distinct orderings of guests and empty seats is $5040 \times 50 = 252000$. This isn't one of the answer choices. So we must assume that some of these orderings are counted as being the same arrangement. maybe if two orderings only differ by a rotation, for example. Or if they only differ by a reflection. Or if they differ by a rotation and/or a reflection ...

I'll let you take it from there.


The method I used to count the number of ways of seating the guests is the following:

First, let´s assume that a round table is nothing but a linear table where the guests seating in the corners are considered to be neighbors.

Now remove the 3 remaining chairs and ask your guests to take a seat (don´t worry about the 3 remaining chairs, we will add them in just a second). This can be made in $7!$ different ways. If each guest is represented by a letter from $A$ to $G$, a possible setting could be $$AEBDCFG,$$ where, as we have said, $A$ and $G$ are considered to be neighbors. But, since the table is round, only $6!$ of these are valid. $$AEBDCFG=GAEBDCF=FGAEBDC=\ldots=EBDCFGA$$

Now take the 3 remaining chairs (each of them will be represented with the letter $Z$) and place each of them in one of the 7 spaces between the guests. This can be done in $\binom{7}{3}$ different ways. Let´s say, for example, that we place them if the following positions $$A\underbrace{\color{\white}{}}_{Z}E\underbrace{\color{\white}{}}_{\color{\white}{a}}B\underbrace{\color{\white}{}}_{\color{\white}{a}}D\underbrace{\color{\white}{}}_{Z}C\underbrace{\color{\white}{}}_{\color{\white}{a}}F\underbrace{\color{\white}{}}_{\color{\white}{a}}G\underbrace{\color{\white}{}}_{Z}\Longrightarrow AZEBDZCFGZ.$$ This way, no remaining chairs are together.

Finally, just multiply the numbers we have just calculated to get a total of $$6!\cdot \binom{7}{3}=25200$$ different ways of seating your guests.


Not the answer you're looking for? Browse other questions tagged or ask your own question.