Five children take part in a tournament. Each one has to play every other one. How many games must they play?

Question

Accepted Answer

To determine the number of games that must be played in a tournament where each of the five children plays every other child exactly once, we can use the concept of combinations. The problem is essentially asking us to find the number of unique pairs of children that can be formed from a group of five.

1. **Explanation for Option 1 (Correct Answer):**

The formula for calculating combinations is given by $ C(n, k) = \frac{n!}{k!(n-k)!} $, where $ n $ is the total number of items to choose from, and $ k $ is the number of items to choose. In this scenario, $ n = 5 $ (the number of children) and $ k = 2 $ (since each game involves a pair of children).

Applying the formula:
   $$
   C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
   $$

Therefore, 10 games must be played, which corresponds to Option 1.

2. **Explanation for why other options are incorrect:**

- **Option 2 (15 games):** This option suggests that 15 games are needed. However, using the combination formula, we have already calculated that only 10 games are necessary. The number 15 would imply that each child plays more than once against some others, which contradicts the condition that each child plays every other child exactly once.

- **Option 3 (17 games):** This option is incorrect because 17 games exceed the number of unique pairs that can be formed from five children. The calculation clearly shows that only 10 unique pairs (games) are possible.

- **Option 4 (20 games):** This option is incorrect as it suggests an even larger number of games than possible. With only 5 children, forming 20 unique pairs is impossible, as it would require each child to play multiple games against the same opponents, which is not allowed under the given conditions.

In conclusion, the correct answer is Option 1, as it accurately reflects the number of unique games that can be played when each child plays every other child exactly once.