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, we will break it down into two parts:

1. Determining how many friends the person should invite to maximize the number of different parties.
2. Calculating how many different combinations of parties can include a specific friend.

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

To maximize the number of different parties, we need to find the value of \( r \) (the number of friends invited) that maximizes the combinations \( \binom{n}{r} \).

Given:
- Total number of friends, \( n = 24 \)

The number of combinations is given by the formula:
\[ \binom{n}{r} = \frac{n!}{r!(n - r)!} \]

The combinations \( \binom{n}{r} \) are maximized when \( r = \frac{n}{2} \).

Calculating:
\[ r = \frac{24}{2} = 12 \]

Thus, the person should invite **12 friends** at a time to maximize the number of different parties.

### Step 2: Calculate the Number of Combinations Including a Specific Friend

Now, we need to find out how many different combinations of parties can include a specific friend.

1. Let's assume we have selected one specific friend.
2. This leaves us with \( 24 - 1 = 23 \) friends.
3. Since we have already selected one friend, we need to select \( 12 - 1 = 11 \) more friends from the remaining 23.

The number of ways to choose these 11 friends from 23 is given by:
\[ \binom{23}{11} = \frac{23!}{11!(23 - 11)!} = \frac{23!}{11! \cdot 12!} \]

Thus, the number of different combinations of parties that include the specific friend is \( \binom{23}{11} \).

### Final Answers:
1. The number of friends to invite at a time is **12**.
2. The number of combinations including a specific friend is \( \binom{23}{11} \).

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