What is the number of seating plans? I am reading a book and there was a topic regarding number of sitting arrangements in a wedding.
The information provided is:

*

*107 people.

*11 tables.

*Each table can accommodate up to 10 people.

*The order in each table does matter but the order of tables does not.

The author finally said that "there are about $11^{107}$ possible seating plans."
Isn't this incorrect since it allows for repetition? My answer would be that the number of seating plans is $110 \choose x_1,...,x_{11}$ such that $x_i = 10$ with $i \in$ {1,...,11}.
Can anyone suggest a correct solution to this problem?
P.S. The book is called "Algorithms to Live By" by Brian Christian and Tom Griffiths.
 A: $11^{107}$ would be the answer if you could assign each person any of the $11$ tables. That means that the tables have no limit on the number of people seated at them, that the seating order around the table does not matter, but the order of the tables does.
When everyone is seated there are $3$ empty seats. It is easiest to introduce $3$ dummy people to fill those seats, and remove them afterwards. So we simply have $110$ people, of which $3$ are identical dummies.
There are $110!$ ways to order all the people. The first $10$ go in order around the first table, the next $10$ around the next table, and so on. Some of these seating arrangements are considered the same, so you now just have to work out how often each arrangement has been overcounted.
The order of the $11$ tables do not matter, so this gives a factor of $11!$ by which we overcounted. Also, the three dummy people that stand in for the empty seats must also be considered the same for a factor of $3!$. This leaves a total of $\frac{110!}{11!3!}$ arrangements.
Note that if we only care about the relative positions of the people seated around a table, then rotation of a table seating does not matter. If that is the case, then you would divide by a further factor of $10^{11}$. Even in this case, the $11^{107}$ estimate is far too low.
A: I'm one of the authors of Algorithms to Live By, and your question came up in my Google alerts. I love that you're engaging so deeply with the book, and wanted to add a couple quick thoughts.
First of all, the wedding that Tom and I are discussing is that of Meghan Bellows and her husband J. D. Luc Peterson. The bride and groom published a delightful paper about how they formalized and solved their wedding seating-chart problem, called "Finding an optimal seating chart" (2012), which I encourage you to check out: https://www.improbable.com/news/2012/Optimal-seating-chart.pdf.
The figure that Tom and I cite comes directly from their paper: "There are a total of $11^{107} = 2.69 × 10^{111}$ combinations for seating guests at the various tables."
I must point out that in both their paper and our book, it is not stated, as you write, that, "The order in each table does matter but the order of tables does not." In fact, the opposite assumptions are made to get the $11^{107}$ figure: namely, that order within a table does not matter, but the tables themselves are distinct.
Honestly, I think your assumptions are more realistic!
But that is not the set of assumptions that Bellows & Peterson use in arriving at their $11^{107}$ figure.
As both you and Jaap discuss, the number of combinations is much greater under your formulation, where order around a table is taken into account. (This may be why Bellows and Peterson did not formulate the problem this way, even though it's intuitive to do so! It is also common at weddings to allow guests to choose their own seats around the table, rather than to have specific place settings. So, that part of the optimization problem is offloaded onto the guests.)
Under this set of assumptions, as you and Jaap note, the figure of $11^{107}$ is too high, because it does not factor into account the maximum capacity of each table. Tom and I mention this in the endnotes for this chapter. However, I seem to recall from talking with Meghan that they relaxed their maximum-capacity constraint on purpose when they ran their mixed-integer linear optimization.
Hope this is helpful to you. I love the discussion on this page!
A: $11^{107}\approx 2.7\times 10^{111}$ would be the number of ways of allocating $107$ distinct people to $11$ distinct tables, where they later find their own seat at the table.  But it allows up to $107$ people at a single table when in fact the number is restricted to $10$ each.
$\dfrac{110!}{{(10!)}^{11}} \approx 1.1\times10^{106}$ would be the number of ways of allocating $110$ distinct people to $11$ distinct tables with $10$ people at each table, again leaving them to find their seats later.  But this is too high as we only have $107$ people, so we need a slightly more detailed calculation depending on the distribution of the three empty places:

*

*If there is an empty place at three tables then the number of ways is ${11 \choose 3}\frac{107!}{{(10!)}^{8}(9!)^{3}}$

*If there are three empty places at one table then the number of ways is ${11 \choose 1}\frac{107!}{{(10!)}^{10}(7!)^{1}}$

*If there are two empty places at one table and one at at another then the number of ways is ${11 \choose 2}{2 \choose 1}\frac{107!}{{(10!)}^{9}(9!)^{1}(8!)^{1}}$
We can add those three possibilities to answer the original question as $$\frac{107!}{{(10!)}^{11}}(165000+7920+99000)\approx 2.3 \times 10^{105
}$$
which is just under a millionth of the original $11^{107}$.
