The Tennis Club Problem Supplementary Notes

1. The Tennis Club Problem Supplementary Notes • A tennis club has 2n members. • We want to pair up the members (by twos) to play singles matches • Question: In how many ways can we pair up all the members of the club?

2. First, from the 2n members we have, choose 2 of them and put them on team 1 • Then, from the remaining 2n-2 members, choose 2 of them and put them on team 2 • Continue this way, at the ith step, choose 2 of the remaining 2n-2(i-1) members and put them on team i

3. How many ways can we perform this process? • By the product principle, the number of ways is

4. Is this the answer to the original question? • No • We have a double counting problem! • Let’s consider an example of the same problem with n = 2 • The tennis club has 4 people and we want to pair them up by twos

5. Suppose the members are a,b,c,d • Consider 2 possible solutions • Solution 1: (a,b) (c,d) • Solution 2: (c,d) (a,b) • They are the same kind of pairing but they are counted twice in the process (double counting) • Although in solution 1 (a,b) are paired up first while in solution 2 (c,d) are paired up first

6. How about n = 3? • Suppose n = 3 and we have members a,b,c,d,e,f • How many “double counting” do we have for the same kind of pairing (a,b) (c,d) (e,f)? • Solution 1: (a,b) (c,d) (e,f) • Solution 2: (a,b) (e,f) (c,d) • Solution 3: (c,d) (a,b) (e,f) • Solution 4: (c,d) (e,f) (a,b) • Solution 5: (e,f) (a,b) (c,d) • Solution 6: (e,f) (c,d) (a,b) • Answer: 6 or 3! • 3! is the way of ordering the 3 pairs

7. Generally, for 2n members, the counting process we introduced would produce n! double counting • n! is the way of ordering the n pairs • Our counting process counts • After eliminating the double counting, we have the solution to the tennis club problem