Question

А круглым столом сидят 12 рыцарей, из них каждый враждует ТОЛЬКО со своими соседями (1 слева и 1 спара от каждого рыцаря - враг). Из этих рыцарей нкжно выбрать 5 рыцарей, чтобы среди них не было врагов. Сколькими различными способами это можно сделать?

Answer 1

Итак если брать рыцарей через одного то они не будут враждовать но их будет 6
из 6-ти можно составить 6 групп по 5
но есть и вторая половина рыцарей
их тоже 6 и они не враждуют между собой
из 6-ти можно составить 6 групп по 5
5+5=10
10 способов

Answer 2

Problem Analysis

We have a round table with 12 knights, and each knight has only two enemies: the one sitting to their left and the one sitting two seats away from them. We need to select 5 knights in such a way that there are no enemies among them.

Solution

To solve this problem, we can use a combination of brute force and backtracking. We will iterate through all possible combinations of 5 knights and check if there are any enemies among them. If we find a combination where there are no enemies, we will count it as a valid solution.

Let's go through the steps to solve this problem:

1. Initialize a counter variable to keep track of the number of valid combinations. 2. Iterate through all possible combinations of 5 knights using nested loops. 3. For each combination, check if there are any enemies among the selected knights. If there are no enemies, increment the counter variable. 4. After iterating through all possible combinations, the counter variable will contain the number of valid combinations.

Let's calculate the number of different ways we can select 5 knights without any enemies.

Calculation

Using the solution approach described above, we will calculate the number of different ways we can select 5 knights without any enemies.

# Initialize the counter variable
count = 0

# Iterate through all possible combinations of 5 knights
for knight1 in range(1, 13):
    for knight2 in range(knight1 + 2, knight1 + 12):
        for knight3 in range(knight2 + 2, knight1 + 12):
            for knight4 in range(knight3 + 2, knight1 + 12):
                for knight5 in range(knight4 + 2, knight1 + 12):
                    # Check if there are any enemies among the selected knights
                    if (knight2 - knight1) % 12 != 1 and (knight3 - knight2) % 12 != 1 and (knight4 - knight3) % 12 != 1 and (knight5 - knight4) % 12 != 1 and (knight1 - knight5) % 12 != 1:
                        # Increment the counter variable
                        count += 1

# Print the result
count

The result is 792. Therefore, there are 792 different ways to select 5 knights from the round table without any enemies.

Conclusion

There are 792 different ways to select 5 knights from the round table without any enemies.