Since there are 4 players in the tournament and each player plays each other player once, then each player plays 3 games.
Since each win earns 5 points and each tie earns 2 points, the possible results for an individual player are:
3 wins, 0 losses, 0 ties: 15 points
2 wins, 0 losses, 1 tie: 12 points
2 wins, 1 loss, 0 ties: 10 points
1 win, 0 losses, 2 ties: 9 points
1 win, 1 loss, 1 tie: 7 points
1 win, 2 losses, 0 ties: 5 points
0 wins, 0 losses, 3 ties: 6 points
0 wins, 1 loss, 2 ties: 4 points
0 wins, 2 losses, 1 tie: 2 points
0 wins, 3 losses, 0 ties: 0 points
In the third table given, Deb has 2 points which means that Deb had 1 tie. If one player has a tie, then another player must also have a tie. But neither 15 points nor 5 points is a possible total to obtain with a tie. Therefore, the third table is not possible.
Similarly, in the fourth table, Ali with 12 points must have had a tie, but none of the other players’ scores allow for have a tie, so the fourth table is not possible.
In the second table, each of Che and Deb must have 2 ties and neither Ali nor Bea can have a tie because of their totals of 10 points each. Since Che and Deb only played each other once, then each of them must have a tie against another player, which is not possible. Therefore, the second table is not possible.
The first table is possible:
Therefore, exactly one of the four given final point distributions is possible.