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Published byEleanor Smith Modified over 9 years ago
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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?
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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
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How many ways can we perform this process? By the product principle, the number of ways is
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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
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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
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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
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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
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