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This question appeared in a discrete mathematics textbook.

A basketball tournament has 16 teams. How many ways are there to match up the teams in 8 pairs?

The way I understand this is that the order that each pair of teams appears within each group of 8 pairs of teams is not relevant (that would be another question). We're just interested in groups consisting of 8 match-ups.

That is, the way I understand the question is that we're interested in all the unique 8-combinations of unordered pairs of teams.

There are 16C2 = 120 possible pairs (match-ups). And there are 120C8 ways of creating unique groups consisting of 8 pairs each.

If we were interested in ordered groups then we would multiply the result by 8!, I suppose.

Does this look fine?

Mariusz Popieluch
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    Some of those $120$ pairs have a team in common, so you can't use two such pairs, so the ${120\choose8}$ is not correct. – Gerry Myerson Apr 26 '23 at 08:32
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    Suppose you had four teams instead of $16$. Can you see the answer would be three? – Gerry Myerson Apr 26 '23 at 08:33
  • Thanks. Perhaps I don't know how such tournaments work. I thought that a team can play more than once in a group -- just that it can't play twice with another identical team. – Mariusz Popieluch Apr 26 '23 at 08:47
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    Maybe it will help if you use balls and bins terminology since this is the standard way this concept is introduced in textbooks. What are the balls and bins in this situation? Are they identical or distinct? – Suraj Apr 26 '23 at 08:49
  • Now that I know I may have misinterpreted the question, I should be fine. How about this: first for each 8-group we choose (pre-choose) one of the teams in each match-up. There would be 16C8 ways to do it. Next we match up the remaining teams without repetition: we can match 8 remaining teams with the first pre-chosen team, 7 with the second pre-chosen team, and so on. So it should be 16C8*8!, I suppose. If this is correct, thanks guys! – Mariusz Popieluch Apr 26 '23 at 09:00
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    Does that work for the case of four teams, Mariusz? – Gerry Myerson Apr 26 '23 at 12:24
  • Wait, indeed it is 3. – Mariusz Popieluch Apr 26 '23 at 13:47
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    @GerryMyerson thank you for your questions! I eventually got to the solution that is presented as method 2 in the answer given by N. F. Taussing. – Mariusz Popieluch Apr 26 '23 at 14:19

1 Answers1

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Method 1: Suppose we list the teams in alphabetical order. There are $15$ ways to select an opponent for the first team in the list. Remove those two teams, which leaves $14$ teams. There are $13$ ways to select an opponent for the first team remaining in the list. Remove those two teams, which leaves $12$ teams. Iterate. There are $$15!! = 15 \cdot 13 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$$ ways to select eight pairs of teams from the $16$ teams in the tournament, where $!!$ means the double factorial.

Method 2: There are $\binom{16}{2}$ ways to select a pair of teams from the $16$ available teams, $\binom{14}{2}$ ways to select a pair of teams from the remaining $14$ teams, $\binom{12}{2}$ ways to select a pair of teams from the remaining $12$ teams, and so forth, giving
$$\binom{16}{2}\binom{14}{2}\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$ ordered selections of eight pairs of teams from the $16$ teams in the tournament. However, the order in which the teams are selected does not matter. Therefore, the number of ways eight pairs of teams may be selected from the $16$ teams in the tournament is $$\frac{1}{8!}\binom{16}{2}\binom{14}{2}\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$

N. F. Taussig
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    Method 3: take an arbitrary ordering of the teams, $(a_1,a_2,\dots,a_{16})$, and interpret it as a pairing, $((a_1,a_2),(a_3,a_4),\dots,(a_{15},a_{16}))$. There are $16!$ orderings of the teams, but there are $8!$ orderings of the eight pairs, all of which give the same pairings, and there are $2$ orderings within each of the eight pairs, so the number of different pairings is $16!/(2^88!)$. – Gerry Myerson Apr 26 '23 at 22:30
  • Thanks for this! Since posting the question realised this approach to be the most clear. We use the elements of the 16-tuples to behave as selectors. Also any unnecessary overcounting is best seen in this model. Of course it is equivalent to the other methods, but to me it seems more direct. – Mariusz Popieluch Oct 17 '23 at 07:59
  • A more general solution of these kind of problems coincides with the ceofficients in Faà di Bruno's formula. I actually learned a lot thinking about the solution to this problem and to the general family of such problems. – Mariusz Popieluch Oct 17 '23 at 08:12