Question

A tennis club has 2n members. We want to pair up the members by twos for singles matches. In how many ways can we pair up all the members of the club? Suppose that in addition to specifying who plays whom, we also determine who serves first for each pairing. Now in how many ways can we specify our pairs? Stein, Cliff L. Discrete Mathematics for Computer Scientists (p. 21). Pearson HE, Inc.. Kindle Edition.'

Answer 1

The first question asks you to take all members and choose any 2.

2n C 2
= 2n! / ((2n - 2)!2!)
= 2n(2n - 1)(2n -2)! / ((2n - 2)!2!)
= 2n(2n - 1) / 2!
= n(2n - 1)
= 2n^2 - n

The second question is essentially saying the “order” of the pair matters (either person 1 serves or person 2 serves):


2n P 2
= 2n! / (2n - 2)!
= 2n(2n - 1)(2n -2)! / (2n - 2)!
= 2n(2n - 1)
= 2n(2n - 1)
= 4n^2 - 2n