Let $M$ be the set of the vertices of a regular hexagon, our Olympiad symbol. How many chains $\emptyset \subset A \subset B \subset C \subset D \subset M$ of six different set, beginning with the empty set and ending with the $M$, are there?
To solve the problem, we need to count the number of chains of sets starting from the empty set and ending with the set \( M \), where \( M \) is the set of vertices of a regular hexagon. The chain must have six different sets.

1. **Identify the number of elements in \( M \)**:
   Since \( M \) is the set of vertices of a regular hexagon, it has 6 elements. Let \( M = \{v_1, v_2, v_3, v_4, v_5, v_6\} \).

2. **Understand the structure of the chain**:
   We need to form a chain of sets \( \emptyset \subset A \subset B \subset C \subset D \subset M \). Each set in the chain must be a subset of \( M \) and must be strictly increasing in size.

3. **Determine the number of maximal chains**:
   A maximal chain in a set of \( n \) elements is a sequence of sets where each set has one more element than the previous set. For a set of 6 elements, the maximal chain would be:
   \[
   \emptyset \subset \{v_1\} \subset \{v_1, v_2\} \subset \{v_1, v_2, v_3\} \subset \{v_1, v_2, v_3, v_4\} \subset \{v_1, v_2, v_3, v_4, v_5\} \subset \{v_1, v_2, v_3, v_4, v_5, v_6\}
   \]
   The number of such maximal chains is \( n! \) for a set of \( n \) elements. Here, \( n = 6 \), so there are \( 6! \) maximal chains.

4. **Adjust for the required chain length**:
   We need chains of length 6, which means we need to remove one set from the maximal chain. There are \( n-1 \) sets of cardinality \( 1 \leq k \leq n-1 \) in the maximal chain. Removing one set from these chains results in sub-maximal chains.

5. **Count the number of sub-maximal chains**:
   When we remove one set from the maximal chain, the resulting sub-maximal chains coalesce two by two. Therefore, the number of such chains is:
   \[
   \frac{(n-1)!}{2} \cdot n!
   \]
   For \( n = 6 \):
   \[
   \frac{(6-1)!}{2} \cdot 6! = \frac{5!}{2} \cdot 6!
   \]

6. **Calculate the final number of chains**:
   \[
   5! = 120
   \]
   \[
   6! = 720
   \]
   \[
   \frac{5!}{2} \cdot 6! = \frac{120}{2} \cdot 720 = 60 \cdot 720 = 43200
   \]

The final answer is \( \boxed{43200} \).