Six distinguishable players are participating in a tennis tournament. Each player plays one match of tennis against every other player. The outcome of each tennis match is a win for one player and a loss for the other players; there are no ties. Suppose that whenever $A$ and $B$ are players in the tournament for which $A$ won (strictly) more matches than $B$ over the course of the tournament, it is also the case that $A$ won the match against $B$ during the tournament. In how many ways could the tournament have gone?
1. **Identify the player with the most wins:**
   - Consider the player who won the most matches and call him \( Q \). According to the problem statement, if \( Q \) won more matches than any other player, then \( Q \) must have won against every other player. Therefore, \( Q \) has 5 wins.

2. **Determine the number of wins for the second-best player:**
   - Now consider the player who won the second most matches. This player must have lost only to \( Q \) and won against the remaining 4 players. Thus, this player has 4 wins.

3. **Continue the pattern for the remaining players:**
   - By continuing this logic, the third-best player must have won against 3 players (losing to \( Q \) and the second-best player), the fourth-best player must have won against 2 players, the fifth-best player must have won against 1 player, and the last player must have won against no one.

4. **Assign the number of wins to each player:**
   - The number of wins for the players are 5, 4, 3, 2, 1, and 0, respectively. Since the players are distinguishable, we need to assign these win counts to each player.

5. **Calculate the number of ways to assign the win counts:**
   - There are 6 players, and we need to assign each of them a unique number of wins from the set \(\{0, 1, 2, 3, 4, 5\}\). The number of ways to do this is \(6!\) (6 factorial).

\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
\]

Conclusion:
\[
\boxed{720}
\]