How many games were played in a basketball game? I came across this question while I was studying for examinations. It looks like this:
"In a basketball game, there are 10 participating teams. If all teams play in a single round robin so that each team can play only once against each team, what is the total number of games played in all?"
The mathematical idea behind the question above seems alien to me. How do you answer the question above? And......what particular topic in mathematics where you deal with these kind of problems?
 A: It is $\displaystyle \binom{10}{2}=45$.
Or we may solve it in the following way:
Each team will play $9$ games. There are $10$ teams, and $10\times 9=90$. But each game is played between two teams. $90$ is actually double counted. So the number of games should be $90\div 2=45$.
A: The first thing you have to notice is that it is a round robin.
Looking at an easier example, with two teams, each team will play against the other once, how many games will there be?
Assuming teams are $A$ and $B$, then there is $1$ game in total: $A$ vs $B$
If there are three, then there is $AvB, AvC, BvC$ so $3$
If there are four, then there is $AvB, AvC, AvD, BvC,BvD, CvD$, so 6.
You might notice a pattern here - assuming the number of teams is $n$, the first team plays $n-1$ games. The second plays $n-2$ games (they actually play $n-1$, but we already counted one of those games, the one against the first team)
Therefore it's the sum from $1 \to n-1$, which is ${(n-1 + 1)(n-1) \over 2} = \frac{n*(n-1)}{2}$ At then, the answer is then $45$
