A bag contains ten balls numbered from 1 to 10.

(b) Show that for any selection of eight numbers, three of the selected
numbers add to 13.

We are tasked with proving that for any selection of eight numbers from the set \( \{1, 2, 3, \dots, 10\} \), there will always be three numbers whose sum is 13.

### Step 1: Partitioning the Set
Let's first partition the set \( \{1, 2, 3, \dots, 10\} \) into pairs of numbers that add up to 13:

\[
(1, 12), (2, 11), (3, 10), (4, 9), (5, 8), (6, 7)
\]

However, the numbers 11 and 12 are not part of the original set \( \{1, 2, 3, \dots, 10\} \), so we discard the pairs \( (1, 12) \) and \( (2, 11) \). This leaves us with the following pairs of numbers that sum to 13:

\[
(3, 10), (4, 9), (5, 8), (6, 7)
\]

### Step 2: Pigeonhole Principle
Now, consider any selection of 8 numbers from the set \( \{1, 2, 3, \dots, 10\} \). The total set contains 10 numbers, and 4 of them form the pairs we identified above.

Each pair contains two numbers that add up to 13. By the **pigeonhole principle**, if we select 8 numbers from the set of 10, at least one of these pairs must be fully selected. Therefore, in any selection of 8 numbers, there will always be at least one pair of numbers whose sum is 13.

### Step 3: Conclusion
Since at least one of the pairs \( (3, 10), (4, 9), (5, 8), (6, 7) \) must be selected in any group of 8 numbers, we are guaranteed to have two numbers that sum to 13. If we take these two numbers and add any other selected number from the group, we will have a triplet of numbers that sum to 13.

Hence, we have shown that for any selection of eight numbers from \( \{1, 2, 3, \dots, 10\} \), three of the selected numbers will add to 13.

---

Would you like more details, or do you have any questions? Here are five related questions to consider:

1. What is the pigeonhole principle, and how is it applied in different mathematical problems?
2. How can this method be extended to different sums or different-sized sets?
3. What are other problems that involve partitioning sets into pairs or groups?
4. How can we generalize this problem for a selection of \( n \) numbers from a set of \( m \) numbers?
5. What happens if we select only 7 numbers instead of 8 in this problem?

**Tip:** When dealing with combinatorics or selection problems, the pigeonhole principle is often a powerful tool for proving the existence of certain conditions.

A bag contains ten balls numbered from 1 to 10. (b) Show that for any selection of eight numbers, three of the selected numbers add to 13.

Explore how the pigeonhole principle can be used to prove that for any selection of eight numbers from a set of ten balls numbered 1 to 10, there will always be three numbers that add up to 13. This combinatorics problem demonstrates the application of set partitioning and the pigeonhole principle, suitable for students in Grades 9-12.

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