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There are $7$ football teams , every team must play just $3$ games , what is the total number of games?

My try follows: ${7 \choose 3}= 35$

MY friend solution

First game : $6$ possible teams

Second games : $5$ possible teams

Third game : $4$ possible teams

So total games: = $6×5×4=120$

Which solution is right?

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    $\begingroup$ What does 'play just' mean? 'play exactly'? Because that means there are $3\times 7\over 2$ games which is impossible. $\endgroup$
    – Shuri2060
    Jul 7, 2017 at 19:51
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    $\begingroup$ Why do you think its 7C3 - that's just choosing 3 teams from the 7? I think I disagree with your friend's solution too. Doesn't that only take into account 1 team's point of view for the number of possible matchups and says nothing about the number of games? $\endgroup$
    – Shuri2060
    Jul 7, 2017 at 19:55
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    $\begingroup$ Ten and a half games. $\endgroup$
    – DanielWainfleet
    Jul 7, 2017 at 19:59
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    $\begingroup$ The explanation for $10.5$ is that each game involves 2 teams. The total number of games is going to be half (the sum of (number of games played by a team) for each team) $\endgroup$
    – Shuri2060
    Jul 7, 2017 at 20:00
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    $\begingroup$ Any answer bigger than $7\times 3$ should raise a red flag in your intuition. Seven teams, three games each: that's 21 games, and that counts each game twice. $\endgroup$
    – G Tony Jacobs
    Jul 7, 2017 at 22:23

2 Answers 2

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Neither answer is correct and I believe this situation is impossible. See the explanation below.

Your answer corresponds to the number of possible unique games that could be played if every game were three-sided. In other words, 7C3 is the number of 3-team groups that could be created from a 7-team league; it has nothing to do with the number of games that might be played.

Your friend's solution corresponds to the number of ways you could have a first, second, and third place team after the season has ended. There are 7 teams that could get first place, 6 remaining teams for second place, and 5 teams remaining that could get third place; hence 7*6*5.

If each of 7 teams played 3 times, that would be a total of 21 games (7*3) if they didn't play each other. I believe your question intends for the teams to play each other. In this case, we would divide by 2 (since team 1 playing team 2 would be a single game, not two games). Therefore we have 21/2=10.5 games, which is clearly impossible.

A couple clarifying examples: if we have 4 teams that each play 2 games, one possible schedule could be 1v2, 1v3, 2v4, 3v4. Each team plays twice and there are 4*2/2=4 games total.

Another impossible scenario might be 5 teams playing 3 games each. An attempt at a schedule might be 1v2, 1v3, 1v4, 2v3, 2v4, 3v5, 4v5, with team 5 missing out on a game while all other teams have already played 3 games.

Thus it the only possible schedules would occur when the product of the number of teams and the number of games each team plays is even; so either the number of teams or the number of games played must be even (or both).

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None of you is correct, and this scenario cannot happen. The total number of games should be $7 \times 3 \over 2$.

If you write down the list of the $N$ matches you will have something like

\begin{array}{| c | c |} \hline \textrm{Team A} & \textrm{Team B}\\ \hline 1 & 2 \\ \hline 3 & 4 \\ \hline 5 & 6 \\ \hline etc. \\ \hline \end{array}

Each team appears 3 times. Therefore the number of teams in this table is $3 \times 7$. And this number is supposed to be equal to $2N$. Then $N=21/2=10.5$ which is not possible.

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  • $\begingroup$ elaborate your answer more please $\endgroup$
    – user373141
    Jul 7, 2017 at 19:56
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    $\begingroup$ @prayersmith ok I added some details $\endgroup$
    – fonfonx
    Jul 7, 2017 at 19:59

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