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A tennis club has 40 members. They host a tournament playing single (one verses one) matches. Every member of the club plays one match with another member of he club, so twenty matches are held. (a) In how many ways can this be arranged?

I was a little confused by the wording of this question in regards to the teams being 'arranged' but I was wondering if it is as simple as the number of ways to choose 20 different match setups, i.e number of ways to choose two different people out of 40. So tried:

C(40,2) and wasn't sure if I needed to multiply by the number of people: so C(40, 2) * 40. I also thought this could be a circular arrangement problem but wasn't sure.. Am I on the right track or way off here?

Newb
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jn025
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  • What’s wanted is the number of different ways to set up $20$ pairs. $\binom{40}2$ gives you only the number of ways to set up one pair. – Brian M. Scott Mar 19 '15 at 04:48
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    See my answer to this question for a pretty thorough discussion of the problem for general $n$ (rather than your $n=20$). – Brian M. Scott Mar 19 '15 at 04:51
  • So by that answer I can just use (220)! / (2^20 20!) ? – jn025 Mar 19 '15 at 05:57
  • That’s the correct answer, but what’s more important is whether you understand the reasoning behind it: the reasoning is likely to be useful in other problems that aren’t just this one with different numbers. – Brian M. Scott Mar 19 '15 at 05:59
  • Thanks and yes I understand the reasoning. Could this be applied to the situation of having teams of two verse teams of two (doubles) by just changing it to 4n pairs? – jn025 Mar 19 '15 at 06:26
  • If the doubles teams are already established, then you can treat each team as a single entity and just count the ways to pair up these entities. Counting the ways to take $4n$ people, pair them into $2n$ doubles teams, and then pair up the teams for the first round of a tournament would be a more complicated problem. – Brian M. Scott Mar 19 '15 at 06:28
  • Okay maybe i don't understand. Assuming that the double teams are already established could you show me what you mean? – jn025 Mar 19 '15 at 06:43
  • If you have $40$ doubles teams, arranging the first round of the tournament for them is just like arranging it for $40$ individuals in a singles tournament, so you get the same result. If you have $80$ people who aren’t yet formed into teams, you can form $40$ doubles teams in $\frac{80!}{2^{40}40!}$ ways, and for each of those ways you can pair up the resulting teams in $\frac{40!}{2^{20}20!}$ ways. – Brian M. Scott Mar 19 '15 at 06:50
  • Even though the teams are already organized the number of people does not change so there would be 20 double teams (as appose to 40) and there would be 10 games instead of 20 – jn025 Mar 19 '15 at 07:13
  • I did change the number of people for my illustration, from $40$ to $80$. But I kept it at $80$ for both parts of my illustration. – Brian M. Scott Mar 19 '15 at 07:14
  • Oh okay that makes sense. So it is still the same answer as having 40 single teams? (I think you already mentioned this, I just want to clarify) – jn025 Mar 19 '15 at 07:34
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    As long as the teams already exist, and we’re just counting the number of ways to pair them up, $40$ teams is $40$ teams, whether they’re teams of $1$, teams of $2$, or teams of $100$. – Brian M. Scott Mar 19 '15 at 07:37

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The Answer should be :

        C(40,2)*C(38,2)*C(36,2)*...*C(2,2)= 40!/(2!)^20

It means we pick 2 out of 40 members then pick 2 out of the 40-2=38 members (the members that we didn't pick last time) and so on.

user224821
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  • With this solution, you're going to be over counting. For example, if you pair up AB first and CD second, you'll count that multiple times because it will be identical to pairing up CD first and AB second. – Duncan Mar 19 '15 at 07:10
  • I'm getting different answers to what you and Brian have said. He suggested to use (2n)! / (2^n * n!) – jn025 Mar 19 '15 at 07:12