2. Andrey, Boris, Vasily, Gennady, and Dmitry played table tennis in pairs such that every two of them played with every other pair exactly once. There were no draws in the tennis matches. It is known that Andrey lost exactly 12 times, and Boris lost exactly 6 times. How many times did Gennady win?

Om vem: Gennady won 8 times.
Solution. The first pair can be formed in $5 \times 4: 2=10$ ways, the second pair can be formed in $3 \times 2: 2=3$ ways. In total, we get $10 \times 3: 2=15$ games. Andrei played in 4 pairs, and they played with 3 pairs. Therefore, Andrei played $4 \times 3=12$ times. According to the problem, he lost 12 times, which means he lost all his games. Together with him, Boris lost 3 times in a pair. Since Boris won 6 times against Andrei when playing with Vasily (2 times), with Gennady, and with Dmitry, the rest of the games he lost, that is, 3 times with Andrei and once each with Vasily (against Gennady and Dmitry), with Gennady, and with Dmitry. Therefore, Gennady lost 3 times with Andrei and once with Boris, a total of 4 times. Thus, he won 8 times.

Criteria. If the solution is incorrect - 0 points.

If the reasoning is correct but there is a computational error - 3 points. If the solution is correct - 7 points.