In the last neighborhood football championship where I live, six teams participated, named $A, B, C, D, E$, and $F$. Each team played exactly one match against each of the others.

$\mathrm{In}$ the championship standings, V indicates the number of wins, $\mathrm{E}$ the number of draws, $\mathrm{D}$ the number of losses, GP the number of goals scored, and GC the number of goals conceded by each team.

(a) How many matches were played?

|  | $\mathbf{V}$ | $\mathbf{E}$ | $\mathbf{D}$ | $\mathbf{G P}$ | $\mathbf{G C}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $A$ | 4 | 1 | 0 | 6 | 2 |
| $B$ | 2 | 1 | 2 | 6 | 6 |
| $C$ | 0 | 3 | 2 | 2 | 6 |
| $D$ | 1 | 1 | $y$ | 3 | 6 |
| $E$ | 0 | 1 | 4 | 1 | 5 |
| $F$ | $x$ | 1 | 0 | $z$ | 3 |

(b) The table is incomplete. Determine the number of wins for team $F$, the number of losses for team $D$, and the number of goals scored by team $F$, represented by $x, y$, and $z$ in the table.
(a) Team $A$ competes with the five teams $B, C, D, E$, and $F$; team $B$, in addition to the match against $A$, already counted, still competes in four matches with teams $C, D, E$, and $F$; team $C$ still competes with teams $D, E$, and $F$; team $D$ still competes with teams $E$ and $F$; and team $E$ still competes with team $F$. In total, we have $5+4+3+2+1=15$ matches played.

Another way to count: We can form groups of two letters and count, remembering that $A B$ and $B A$, for example, are the same match: $A B, A C, A D, A E, A F, B C, B D, B E, B F, C D, C E, C F, D E, D F$, and $E F$ gives 15 matches.

Another way to count: Each of the six teams played exactly one match with each of the other five. Therefore, a total of $\frac{1}{2}(6 \times 5)=15$ matches were played.

(b) Each team played five matches, so the sum of the number of wins, draws, and losses for each team is equal to five. Thus, we have $x+1+0=5$, so $x=4$, and we have $1+1+y=5$, so $y=3$.

Another way to calculate $x$ and $y$: We know the total number of draws (which always involve two teams), given by $1+1+3+1+1+1=8$. Therefore, the number of wins (or losses) is equal to the number of matches, 15, minus half the number of draws, 4, or the total number of wins (or losses) is $15-4=11$. Thus, $4+2+1+x=11$, so $x=4$, and $2+2+y+4=11$, so $y=3$. Finally, the total number of goals scored in the tournament is equal to the total number of goals conceded, which is $2+6+6+6+5+3=28$. Thus,

$$
28=6+6+2+3+1+z
$$

that is, $z=10$.

In summary, the number $x$ of wins for team $F$ is 4, the number $y$ of losses for team $D$ is 3, and the number $z$ of goals scored by team $F$ is 10.