A tennis club has $2n$ members. We want to pair up the members by twos for singles match. How many ways may we pair up all the members of the club?

Question

This question is from Kenneth P. Bogart's book "Combinatorics through guided discovery" which can be freely downloaded at

https://math.dartmouth.edu/news-resources/electronic/kpbogart/FDL-bogart/kpbogart-FDL.tar.gz

The question sounds very simple and, I thought the solution was $C_{(2n, 2)}$ which after some simple algebra yields $C_{(2n,2)} = n(2n - 1)$.

For a simple (and manually verifiable) case of 10 members, thus $n = 5$, it leads to the answer 45 ways which is the same result as the one obtained from $C_{(10,2)}$.

The answer in the book bears no resemblence to that. Specifically, quoting from the book

12. A tennis club has $2n$ members. We want to pair up the members by twos for singles matches.

(a) In how many ways may we pair up all the members of the club? (Hint: consider the cases of 2, 4, and 6 members.)

Solution:

Suppose we list the people in the club in some way (perhaps in alphabetical order), and keep that list for the remainder of the problem.

Take the first person from the list and pair that person with any of the $2n − 1$ remaining people. Now take the next unpaired person from the list and pair that person with any of the remaining $2n − 3$ unpaired people. Continuing in this way, once k pairs have been selected, take the next unpaired person from the list and pair that person with any of the remaining $2n − 2k − 1$ unpaired people.

Every pairing can arise in this way, and no pairing can arise twice in this process.

Thus the number of outcomes is $$\prod_{i=0}^{n-1}2n − 2i − 1$$.

This is the product of the odd numbers between 1 and $2n − 1$. It is also the product of every second term of $(2n)!$. Notice that the other n terms of $(2n)!$ are even. In fact, if we double each term of n!, we would get the missing terms of (2n)!. Thus $$(2n)! = 2^n n! \sum_{i=0}^{n-1} 2n - 2i - 1$$

Therefore the number of tennis pairings is $$\frac{(2n)!}{2^n\ n!}$$.

end quote from the book

I do not see how the number of outcomes ends up as the product $$\prod_{i=0}^{n-1}2n − 2i − 1$$ I see a sum of elements of disjoint sets not a product.

for $2n = 10$, I see $$C_{10,2} = 45$$ his solution yields $$\frac{(10)!}{2^5\ 5!} = 945$$

That's 900 additional pairs that I do not see. What pairs are they counting that I am not seeing ?

Thank you. (excuse my formatting, I'm learning MathJax/LaTex on the fly)

Answer 1

Your answer is 45 is correct for the case your aim is to conduct a single match. It is a matter of choosing two players from among $2n$ players. But the question is to organize many matches such that all players get to play a match. So $n$ matches all involving a new pair will cover all the $2n$ players. That is what is calculated in the book. That is how these $n$ matches could be played. Here is a partial listing with 6 players A ,B,C,D,E, F

(1) A vs B, C vs D, E vs F

(2) A vs B, C vs F, E vs D

(3) A vs C, B vs D, E vs F

etc

Answer 2

$C_{(n,k)}$ means that you choose a subset $k$, from a set of $n$.

When you perform, say, $C_{(2n,2)}$ means that you are choosing only 2 members out of a total $2n$, and not pairing up the members as you thought you were.

Here's a hint : Consider person 1 - How many people can he choose from?

How many left? $2n-2$.

So from how many can person 2 choose?

And so on...

Answer 3

Consider writing a list of numbers from 1 to 2n as a coding of the n matches: 3 1 4 5 9 2 6 7 ... would mean 3 and 1 play a match, as to 4 and 5, 9 and 2 6 and 7... There are (2n)! ways of writing this list.

But there are also multiplicities---multiple ways we can write down the same set of matches. If we had the $2n$ people playing $n$ matches, we could have written the matches down in any of $n!$ order ways. And for each of the matches, we could have written the two participants down in 2 ways (if they are $i$ and $j$, either $i j$ or $j i$), so we could have ordered all the participants in $2^n$ ways.

Therefore each of the matchings could be written down in $n!2^n$ ways. As a consequence, the number of matchings is $(2n)!/(n!2^n)$.