1. In a tennis tournament, there are $2^{n}+4$ schoolchildren participating, where $n-$ is a natural number greater than 4. A win earns 1 point, and a loss earns 0 points. Before each round, pairs are formed by drawing lots from participants who have the same number of points (those who do not find a pair skip the round). The tournament ends as soon as a sole leader is determined. How many schoolchildren will have 3 points after the tournament ends?
Answer: $C_{n}^{3}+1$.

Solution. First, let's prove two statements.

1) In a tournament with $2^{n}$ participants, there will never be a player without a pair, and by the end of the tournament, for any $k \in\{0, \ldots, n\}$, $C_{n}^{k}$ students will have $k$ points. We will use induction on $n$. For $n=1$, everything is obvious. Suppose that for some $n$ both statements are true. Then, in a tournament with $2^{n+1}$ participants, after the first round, $2^{n}$ students will have 0 points and $2^{n}$ students will have 1 point. We will continue to conduct the draw so that participants from these groups only meet each other. By the inductive hypothesis, this can always be done, and such an assumption will obviously not affect the distribution of points. In other words, we will further conduct two independent tournaments with $2^{n}$ participants. By the inductive hypothesis, there will be no players without a pair in these tournaments. Moreover, in the first tournament, $C_{n}^{k}$ students will have $k$ points, and in the second tournament, $C_{n}^{k-1}$ students will have $k$ points, since all participants started with one point. In the end, $C_{n}^{k}+C_{n}^{k-1}=C_{n+1}^{k}$ students will have $k$ points (we used the main identity of Pascal's triangle). Thus, the inductive step is completed.
2) The tournament can be reduced to two independent tournaments in groups of $2^{n}$ and 4 people. Indeed, in each round, we will only pair participants from the same group. By 1), all participants in the first group can be paired. Therefore, if someone is left without a pair in the overall tournament, we can assume that this participant is from the second group. At the end of the tournament, the results of the two groups are simply combined.

Let's calculate the total number of participants who scored 3 points in the tournament. By 1), in the first group, there will be $C_{n}^{3}$ such participants. The dynamics of the distribution of points among participants from the second group are as follows:

$$
4 \rightarrow(2,2) \rightarrow(1,2,1) \rightarrow(1,1,2) \rightarrow(1,1,1,1)
$$

(the left number represents the number of underdogs, the right number represents the number of leaders). After the fourth round, participants from the second group have no one to play with, so at the end of the tournament, one participant will have 3 points. Thus, the answer is the sum of $C_{n}^{3}$ and 1.