Ali, Bea, Che, and Deb compete in a checkers tournament. Each player plays each other player exactly once. At the end of each game, either the two players tie or one player wins and the other player loses. A player earns 5 points for a win, 0 points for a loss, and 2 points for a tie. Exactly how many of the following final point distributions are possible?

| Player | Points | Player | Points | Player | Points | Player | Points |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Ali | 15 | Ali | 10 | Ali | 15 | Ali | 12 |
| Bea | 7 | Bea | 10 | Bea | 5 | Bea | 10 |
| Che | 4 | Che | 4 | Che | 5 | Che | 5 |
| Deb | 2 | Deb | 4 | Deb | 2 | Deb | 0 |

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
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:

| Result | Ali | Bea | Che | Deb |
| :---: | :---: | :---: | :---: | :---: |
| Ali wins against Bea | 5 points | 0 points |  |  |
| Ali wins against Che  Ali wins against Deb | 5 points |  | 0 points |  |
| Bea ties Che  Beans against Deb  Che ties Deb |  | 2 points | 2 points | 0 points |
| TOTAL | 15 points | 7 points | 4 points | 2 points |

Therefore, exactly one of the four given final point distributions is possible.

ANsWer: (B)