Forming Dancing Pairs Among Couples: A Combinatorial Approach

How many ways can n couples form dancing pairs either with their own spouse or by exchanging spouses? In this article, we explore a combinatorial approach to solve this intriguing problem.

Breakdown of the Problem

The problem involves n couples, making a total of 2n individuals. Each individual can either dance with their own spouse or with another couple's spouse, leading to a series of potential pairings.

Pair Formation

Each of the n couples has two choices:

They can dance with their own spouse. They can split and pair with someone else's spouse.

Counting the Pairings

To accurately count the number of valid pairings, we can use a combinatorial method and the principle of inclusion-exclusion.

Approach Using Inclusion-Exclusion

We start by considering the total number of pairings without any restrictions. For each individual, there are initially 2n - 1 possible partners, then 2n - 3, and so on.

Total Pairings

However, to respect the couple pairings, we need to ensure that:

Each couple either dances together or not at all. Valid pairings must preserve the pairing constraints.

Recursive Formula

A recursive formula can be derived using the principle of inclusion-exclusion:

Recursive Formula

The number of ways to pair n couples, denoted as Pn, can be defined as:

Pn Pn-1 2n-1 cdot Pn-2

Where Pn is the number of ways to pair n couples, and Pn-1 and Pn-2 are the base cases.

Base Cases

The base cases are as follows:

P0 1: With no couples, there is one way to do nothing. P1 1: With one couple, they can only dance with each other.

Final Calculation

Using the recursive formula, we can compute Pn for any n. Here are a few examples to illustrate the process:

Calculate P2

P2 P1 2 cdot P0 1 2 cdot 1 3

Calculate P3

P3 P2 4 cdot P1 3 4 cdot 1 7

We can continue this process to find Pn for any n.

Conclusion

Thus, the number of ways n couples can form dancing pairs is given by the recurrence relation:

Pn Pn-1 2n-1 cdot Pn-2, with base cases P0 1 and P1 1. This can be computed for any n to find the total number of valid dancing pairs.