Question

Двенадцать шахматистов договорились сыграть несколько партий так, чтобы любые два шахматиста играли друг с другом не более одного раза. К некоторому моменту было сыграно 44 партии, причём каждый шахматист сыграл чётное количество партий, а один всё это время был болен, наконец-то поправился и теперь смог принять участие в турнире. Сколькими способами можно провести ещё несколько партий так, чтобы каждый шахматист сыграл нечётное число партий? Способы, отличающиеся только порядком сыгранных партий, считаются одинаковыми.

Answer 1

Problem Analysis

We have 12 chess players who want to play several games in such a way that no two players play against each other more than once. At a certain point, 44 games have already been played, with each player playing an even number of games. One player was sick but has now recovered and can participate in the tournament. We need to determine how many ways we can play additional games so that each player plays an odd number of games.

Solution

To solve this problem, we can use the concept of graph theory. We can represent the players as nodes in a graph, and the games they have played as edges between the nodes. Since each player has played an even number of games, the degree of each node (player) in the graph is even.

To ensure that each player plays an odd number of games, we need to add additional edges to the graph. However, we need to consider that the graph should remain connected, and no player should play against another player more than once.

To add the additional edges, we can start by connecting the recovered player (who was sick) to an even number of other players. This will ensure that the recovered player's degree becomes odd. We can then continue adding edges between the remaining players in pairs until each player's degree becomes odd.

Let's calculate the number of ways we can add the additional edges.

Calculation

Since the recovered player needs to be connected to an even number of players, we have the following possibilities: - Connect the recovered player to 0 players (0 ways) - Connect the recovered player to 2 players (1 way) - Connect the recovered player to 4 players (C(10, 4) ways) - Connect the recovered player to 6 players (C(10, 6) ways) - Connect the recovered player to 8 players (C(10, 8) ways) - Connect the recovered player to 10 players (1 way)

After connecting the recovered player, we need to pair up the remaining players and connect them to each other. Since there are 10 remaining players, we need to calculate the number of ways to pair them up.

The number of ways to pair up 10 players can be calculated using the formula for derangements (also known as subfactorial): D(10) = 10! * (1/0! - 1/1! + 1/2! - 1/3! + ... + (-1)^10/10!)

Finally, we can calculate the total number of ways to add the additional edges by multiplying the number of ways to connect the recovered player and the number of ways to pair up the remaining players.

Let's calculate the result.

Calculation Steps

1. Calculate the number of ways to connect the recovered player: - Connect the recovered player to 0 players: 0 ways - Connect the recovered player to 2 players: 1 way - Connect the recovered player to 4 players: C(10, 4) ways - Connect the recovered player to 6 players: C(10, 6) ways - Connect the recovered player to 8 players: C(10, 8) ways - Connect the recovered player to 10 players: 1 way

2. Calculate the number of ways to pair up the remaining players: - Calculate the derangement of 10 players: D(10)

3. Calculate the total number of ways to add the additional edges: - Multiply the number of ways to connect the recovered player and the number of ways to pair up the remaining players.

Let's calculate the result.

Calculation Result

The total number of ways to add the additional edges so that each player plays an odd number of games is (0 ways + 1 way + C(10, 4) ways + C(10, 6) ways + C(10, 8) ways + 1 way) * D(10).

Please note that the calculation of combinations (C) and derangements (D) involves factorials and can be quite complex. To get the exact numerical result, you may need to use a calculator or a programming language with support for large numbers.

I hope this explanation helps! Let me know if you have any further questions.