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?

  • 6
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ Ten and a half games. $\endgroup$ Jul 7, 2017 at 19:59
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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. $\endgroup$ Jul 7, 2017 at 22:23

2 Answers 2


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).


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.

  • $\begingroup$ elaborate your answer more please $\endgroup$
    – user373141
    Jul 7, 2017 at 19:56
  • 1
    $\begingroup$ @prayersmith ok I added some details $\endgroup$
    – fonfonx
    Jul 7, 2017 at 19:59

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