While watching a sports league the following question came to my mind. Consider a sports league which has $8$ teams. Each team plays with another team $2$ times. Hence, the total matches played will be $8C_2\times2 =56$. Suppose all the teams have played $10$ matches. How many matches are remaining in league to be played?
My approach:
Since each team plays a total of $14$ matches, if each team had played $10$ matches, they have played half of their matches. Hence $28$ matches are already played. However, the problem occurs how to tackle remaining $3$ matches? I mean how can I use this to solve problem. Is there any alternative way?
You are correct that each team plays a total of $14$ matches and that the total number of matches played is $2\binom{8}{2} = 56$.
Another way to see that the total number of matches played is $56$ is to observe that if we multiply the number of teams by the number of matches each team played, we will have counted the number of matches twice since each match involves two teams. Therefore, the total number of matches played is $$\frac{8 \cdot 14}{2} = 8 \cdot 7 = 56$$ After each team has played $10$ matches, they have played a total of $$\frac{8 \cdot 10}{2} = 8 \cdot 5 = 40$$ of those $56$ matches, so the number of matches remaining to be played is $56 - 40 = 16$.
Alternatively, since each team has played $10$ of its $14$ matches, each of the eight teams has $14 - 10 = 4$ matches remaining. Consequently, the number of matches that remain is $$\frac{8 \cdot 4}{2} = 8 \cdot 2 = 16$$