A person wants to hold as many different parties as he can out of 24 friends, each party consisting of the same number. How many should he invite at a time? In how many of these would the same man be found?

Answer

To solve the problem step by step, we will break it down into two parts: determining how many friends the person should invite at a time and finding out how many different parties can be formed with the same person included.

### Step 1: Determine the Maximum Number of Friends to Invite

1. **Understanding the Problem**: The person wants to hold parties with the same number of friends. To maximize the number of different parties, we need to find the optimal number of friends to invite at a time.

2. **Choosing Friends**: The number of ways to choose `r` friends from 24 friends is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).

3. **Finding the Maximum**: The maximum number of friends that can be invited at a time occurs when `r` is half of the total number of friends. Since there are 24 friends, the maximum number of friends to invite at a time is \( r = \frac{24}{2} = 12 \).

4. **Conclusion for Part 1**: The person should invite **12 friends** at a time.

### Step 2: Determine How Many Parties Include the Same Person

1. **Selecting One Person**: Let's say the person wants to find out how many parties include a specific friend (let's call this friend A).

2. **Remaining Friends**: If A is already invited, there are 23 friends left (since one person is already included).

3. **Choosing the Remaining Friends**: Out of the remaining 23 friends, we need to choose 11 more friends to make a total of 12 friends in the party (including A). The number of ways to choose 11 friends from 23 is given by \( C(23, 11) \).

4. **Conclusion for Part 2**: The number of different parties that include the same person is \( C(23, 11) \).

### Final Answers:
- **Part 1**: The person should invite **12 friends** at a time.
- **Part 2**: The number of parties that include the same person is \( C(23, 11) \).