# The Tennis Club Problem Supplementary Notes - PowerPoint PPT Presentation

1 / 7

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?.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

The Tennis Club Problem Supplementary Notes

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

### 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?

• 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

• How many ways can we perform this process?

• By the product principle, the number of ways is

• 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

• 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

• 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)