Jonathan is leading his PHINS group across campus when he hears that Andrew got lost in the basement of DRL. To ensure no one else gets lost in the depths of DRL, Jonathan insists that his students travel using the “buddy” system, meaning that his students must pair up before entering DRL. Jonathan’s group happens to consist of 864 distinct students. How many ways can Jonathan pair up students in his group to search for Andrew? (Note that Andrew is not part of the 864 students, the order of the students within a pair does not matter, and the ordering of the pairs doesn’t matter either).

We are tasked with finding how many ways Jonathan can pair up 864 distinct students such that the order within each pair does not matter, and the ordering of the pairs themselves does not matter. This is a combinatorial problem.

### Step 1: Number of ways to form pairs

Jonathan has 864 students, and they need to be paired up in groups of 2. To begin, let's think about the number of ways to divide the 864 students into pairs.

#### Total ways to arrange students:
First, consider the number of ways to arrange all 864 students in a line. This is given by \( 864! \) (factorial of 864).

#### Divide by the ways to order within a pair:
Each pair of students can be arranged in two ways (since the order within the pair doesn't matter). Since there are 432 pairs (since \( \frac{864}{2} = 432 \)), we divide by \( 2^{432} \) to account for the fact that within each pair, the order does not matter.

#### Divide by the ways to order the pairs:
The order of the pairs themselves also doesn't matter, so we divide by \( 432! \) to account for the number of ways to arrange the 432 pairs.

Thus, the total number of ways to pair up the 864 students is:

\[
\frac{864!}{2^{432} \times 432!}
\]

### Step 2: Final answer

The formula we derived:

\[
\frac{864!}{2^{432} \times 432!}
\]

represents the total number of ways Jonathan can pair up the students. This is the final expression, and while it's difficult to compute by hand, it is a correct and simplified representation of the solution.

Would you like more details on how to compute this, or do you have any further questions?

### Related questions:
1. How would the formula change if Jonathan had 866 students instead of 864?
2. What if the pairs were ordered (i.e., the order of students in a pair mattered)?
3. What if Jonathan wanted to group the students into triplets instead of pairs?
4. How does this problem change if a few students are already paired up, and the remaining need to be paired?
5. What is the probability that a specific student is paired with a specific other student?

#### Tip: When faced with factorials and large combinatorial numbers, using logarithms or approximation techniques (like Stirling's approximation) can help simplify computations.

Discover the combinatorics behind pairing 864 students into 432 pairs using the formula (864! / (2^432 × 432!)). This advanced problem involves factorials, combinatorial arrangements, and counting techniques. Suitable for advanced math students or those studying combinatorics.

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