Set has the property that for any two triplets and in , we have , and also, we know that at most one of the equalities , , holds. Maximize .
Solution
To solve this problem, we need to maximize the size of the set under the given constraints. Let's break down the problem step by step.
1. Understanding the Constraints:
- Each triplet must satisfy .
- For any two distinct triplets and in , at most one of the equalities , , or can hold.
2. Analyzing the Triplet Structure:
- Since , each triplet is strictly increasing.
- We need to ensure that no two triplets share more than one coordinate.
3. **Constructing the Set :**
- Consider the set of all possible triplets where .
- The number of such triplets is given by the binomial coefficient , which counts the number of ways to choose 3 distinct elements from elements in increasing order.
4. Ensuring the Constraint on Equalities:
- We need to ensure that no two triplets share more than one coordinate.
- One way to achieve this is to select triplets such that no two triplets have the same first, second, or third element.
5. **Maximizing the Size of :**
- To maximize , we need to find the largest possible subset of triplets that satisfies the constraints.
- A useful approach is to consider the combinatorial structure and use the pigeonhole principle to ensure that no two triplets share more than one coordinate.
6. **Constructing a Valid Set :**
- One possible construction is to choose triplets such that each element appears in a unique position across different triplets.
- For example, if , we can choose the triplets , , , ensuring that no two triplets share more than one coordinate.
7. Generalizing the Construction:
- For a general , we can construct a set by carefully selecting triplets to avoid sharing more than one coordinate.
- The maximum size of such a set can be shown to be , as this is the largest number of triplets that can be chosen without violating the constraints.