There were $107$ people at the wedding and $11$ tables, which could accommodate $10$ people each. 
There were $107$ people at the wedding and $11$ tables, which could accommodate $10$ people each. This means there were about $11^{107}$ possible seating plans.

I read that in a book but didn't really understand the solution. I mean the solution doesn't consider the fact that each table can have a max of $10$ people and it actually doesn't care about the capacity of the table. Can anyone explain it to me?
 A: The simplest way to read the problem is to consider all the seats distinct, so there are $110$ of them.  In that case the first person has $110$ choices of seat, the next has $109$, the next has $108$ and so on.  This gives $\frac {110!}{3!}$ possible seating plans.  As in my comment, you may want to consider rotating people around a table the same seating plan.  Each table has $11$ rotations, so you should divide by $11^{10}$.  Maybe you want to consider swapping groups of people among the tables, so the tables are indistinguishable.  Then you should divide by $11!$.  You need to specify what makes a seating plan unique.
A: $11^{107} \approx 2.7 \times 10^{111}$ is the number of ways you allocate each person to a table, not a seat.  But you could end up with more than $10$ people to a table in that count.
$\frac{110!}{10!^{11}}\approx 1.1 \times 10^{106}$ would be the way of allocating $110$ people to tables with $10$ at each table.  But there will be $110-107 =3$ empty spaces, so I think that number is perhaps $\frac{110\times 109 \times 108}{{11 \choose 1} \times 10\times 9 \times 8+{11 \choose 2}{2\choose 1}\times 10^2\times 9 +{11 \choose3}10^3 } \approx 4.76$ times too high so perhaps about $2.3 \times 10^{105}$ is the true figure.
So $11^{107}$ is too big on this basis, perhaps out by a factor of a million, but this is not that dramatic compared with other possible changes to the calculation: you would get a smaller number if individual tables did not matter, dividing by $11!\approx 4 \times 10^7$, while if you wanted to allocate seats as well as tables too you would get Ross Millikan's much higher numbers.
A: Let $a(n,t)$ be the number of ways to assign $n$ distinguishable people to $t$ indistinguishable tables, each with capacity $c$.  Without the capacity, this is a Stirling number of the second kind.  By conditioning on the number $k$ of people at the table that contains the first person, we obtain
$$a(n,t) =
\begin{cases}
1 &\text{if $n = 0$} \\
0 &\text{if $t = 0$}\\
\sum_{k=1}^{\min(c,n)} \binom{n-1}{k-1} a(n-k,t-1) &\text{otherwise}
\end{cases}
$$
For $c=10$, we have $a(107,11)=5.8152\times 10^{97}$.
Let $b(n,t)$ be the number of ways to seat $n$ distinguishable people around $t$ indistinguishable tables, each with capacity $c$.  Without the capacity, this is a Stirling number of the first kind.  By conditioning on the number $k$ of people at the table that contains the first person, we obtain
$$b(n,t) =
\begin{cases}
1 &\text{if $n = 0$} \\
0 &\text{if $t = 0$}\\
\sum_{k=1}^{\min(c,n)} (k-1)! \binom{n-1}{k-1} a(n-k,t-1) &\text{otherwise}
\end{cases}
$$
For $c=10$, we have $b(107,11)=1.2132\times 10^{156}$.  If rotations are not equivalent, multiply each summand by $k!$ instead of $(k-1)!$.
For both $a(n,t)$ and $b(n,t)$, if the tables are instead distinguishable, multiply the final results by $t!$.
