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