Let $F(n)$ be the number of ways of seating $n$ people. We could start by carefully listing the ways, for a few small values of $n$. The listing should be almost explicit. We find that $F(1)=1$, $F(2)=2$, $F(3)=6$, and (perhaps) that $F(4)=24$. Despite the small amount of evidence, the conjecture $F(n)=n!$ is tempting.
We prove the conjecture, in principle by induction. Suppose that we know that for a specific $k$, $F(k)=k!$. Now George, the $(k+1)$-th person, comes along, late as usual. For every possible seating of the $k$ people, we can place George at the table of one of the $k$ people, and immediately to the right of that person ($k$ choices), or we can place George at a table by himself. Thus
$$F(k+1)=kF(k)+F(k)=(k+1)F(k)=(k+1)!.$$
Since $F(1)=1$, we conclude that $F(n)=n!$ for all $n$.
Another way: We can use fancier language. For example, note that every permutation of the set $\{1,2,\dots,n\}$ can be expressed uniquely as a product of disjoint cycles. We explicitly include any $1$-cycles. The order in which the product is taken does not matter, since the cycles are disjoint.
Every product of disjoint cycles corresponds to a unique circular seating, and vice-versa. (The people in a cycle determine a table, and their cyclic order determines the order of seating at that table.) This gives an explicit bijection between the permutations of $\{1,2,\dots,n\}$ and the seatings. Thus the number of seatings is equal to $n!$, the number of permutations.
There are $100!$ ways to line up the people. Take the first $5$ and seat them around the first table in order. Do the same for the second $5$ and the second table. Continue until you’ve seated all $100$. If the tables are individually identified, and each of the $20$ tables has a designated head seat, each of these $100!$ seatings counts as a different arrangement. If the tables are individually identified, but seatings that are equivalent under a rotation of the table are not considered distinct, then each seating at each table has been counted $5$ times, once for each possible location of a head seat, and there are therefore only $\dfrac{100!}{5^{20}}$ distinct arrangements.
You appear to be assuming, however, that the tables are not individually identified and that rotationally equivalent arrangements at a table are not distinguished. In that case each distinct arrangement has been counted $20!$ times in the figure $\dfrac{100!}{5^{20}}$, once for each of the $20!$ permutations of the tables, so there are only $\dfrac{100!}{5^{20}20!}$ distinct arrangements. This is a bit over $4\times10^{125}$.
Best Answer
Forget about the tables.
There are $\binom{6}{3}=20$ ways to choose $3$ people. This double-counts the number of ways to divide the $6$ people into two groups of $3$, for the choice of A, B, C leads to the same splitting as the choice D, E, F. So the number of splittings is $10$.
By contrast, there are $\binom{4}{2}$ ways to split the group into $2$ parts, one with $4$ and the other with $2$. No splitting gets double-counted.
The circular permutation parts are dealt with in the usual way.