a) A championship will have 7 competitors. Each of them will play exactly one game against all the others. What is the total number of games in the championship?

b) A championship will have $n$ competitors and each of them will play exactly one game against all the others. Verify that the total number of games is $\frac{n \cdot(n-1)}{2}$.

c) A competitor $A$ is called a master of the championship if every other player either lost to $A$ or lost to someone who lost to $A$. Verify that every championship in which each player plays exactly once against all the others and there are no ties, always admits a master.

d) João and Maria were in a tournament where each athlete was supposed to play against the other exactly once. However, Maria had a health issue and abandoned the tournament after playing 10 games, and João had to leave after his first game. At the end of the day, a total of 55 games were counted. Was the game João played against Maria?
Solution

a) Let $\{a, b, c, d, e, f, g\}$ be the players. Player $a$ will participate in the 6 games associated with the pairs: $(a, b), (a, c), (a, d), (a, e), (a, f)$, and $(a, g)$. Initially, since we have already counted the game associated with $(a, b)$, to count the games for player $b$, we only need to count the games associated with the 5 pairs $(b, c), (b, d), (b, e), (b, f)$, and $(b, g)$. By subtracting the already counted games and repeating this analysis for the next players, we obtain the following number of games: 4, 3, 2, and 1. Therefore, the total number of games is

$$
6+5+4+3+2+1=\frac{6 \cdot 7}{2}=21
$$

b) Repeating the count from the previous item, if $\left\{j_{1}, j_{2}, \ldots, j_{n}\right\}$ are the players, the games that $j_{1}$ will participate in are associated with the $n-1$ pairs:

$$
\left(j_{1}, j_{2}\right), \left(j_{1}, j_{3}\right), \left(j_{1}, j_{4}\right), \ldots, \left(j_{1}, j_{n}\right)
$$

In general, after listing the pairs of players $j_{1}, j_{2}, \ldots, j_{k-1}$ without repetition, we need to count for player $j_{k}$ the $n-k-1$ games associated with the pairs:

$$
\left(j_{k}, j_{k+1}\right), \left(j_{k}, j_{k+2}\right), \left(j_{k}, j_{k+3}\right), \ldots, \left(j_{k}, j_{n}\right)
$$

Therefore, the total number of games is

$$
1+2+3+\ldots+n=\frac{n(n-1)}{2}
$$

Note that this is exactly the number of pairs $\left(j_{l}, j_{m}\right)$ with $l < m$ and could be obtained with the following alternative count: There are $n$ options for the first element of the pair and $n-1$ for the second element, producing $n(n-1)$ pairs of the form $\left(j_{a}, j_{b}\right)$ with $a \neq b$. However, the two pairs $\left(j_{a}, j_{b}\right)$ and $\left(j_{b}, j_{a}\right)$ were counted, and only one of them has the player with the smaller index in the first position. Thus, it suffices to divide the previous count by 2, obtaining $\frac{n(n-1)}{2}$.

c) Let $A$ be the player with the most wins in the tournament. Consider any other competitor $B$. Two situations can occur: