2. In a round-robin chess tournament, 7 players participated. How many games did they play?
2. Let's calculate the total number of games played in the tournament. Each player played 6 games - in total: $6 \times 7=42$. Notice that in this way, we counted each game twice (since two players participate in each game!), so the total number of games played in the tournament is $42: 2=21$. We can also solve this using a table:

| № | Participants | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | Ivanov | 口口 |  | 1 |  |  |  |  |
| 2 | Svirin |  | 口口 |  |  |  |  |  |
| 3 | Sherstnev | 0 |  | 口■ | \| |  |  |  |
| 4 | Sudo |  |  | $\mid$ | $\square \square$ |  |  |  |
| 5 | Makarov |  |  |  |  | $\square \square$ |  |  |
| 6 | Romanov |  |  |  |  |  | $\square \square$ |  |
| 7 | Novikov |  |  |  |  |  |  | $\square \square$ |

Tournaments $\quad 72$

From the table, it becomes clear that the total number of games played is $7+6+5+\ldots+1$ - the number of cells located above the diagonal.