In a tennis tournament in which every pair has to play a match with every other pair, $10$ players are playing. Find the number of games played.
Attempt: Number of ways of selecting $2$ players out of $10$ for forming a pair is same as arranging them in a row and then dividing them into groups of two
$$=\frac{10!}{(2!)^55!}=945$$
Total number of possible pairs =$945.$
These $945$ pairs have to play a match with every other possible pair.Since a pair can't play a match against itself and any other pair which has one of its member.
A pair can play only against the pairs formed by other $8$ players i.e.
$=\frac{8!}{(2!)^44!}=105$ other pairs.
So the no of games played should be $105*5$ but the text mentions the answer as $105*6$.
Where am I going wrong? What's the other way to solve this?
Best Answer
Here's one approach:
There are ${{10}\choose {2}}=45$ pairs.
For each of these pairs there are ${{8}\choose {2}}=28$ pairs that they must play.
So there would be $45\cdot 28$ matches played, except this double counts each match (once for each involved pair).
So there are $\frac{45\cdot 28}{2}$ matches to play.