# Seven guests turned up for dinner at a round table with 10 seats. [closed]

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:

604800.

30240.

25200.

5040.

• 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? 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.