Consider six players $P_1, P_2, P_3, P_4, P_5$ and $P_6$. A team consists of two players.
(Thus, there are $15$ distinct teams.) Two teams play a match exactly once if there is no common player. For example, team $\{P_1, P_2\}$ can not play with $\{P_2, P_3\}$ but will play with $\{P_4, P_5\}$. Then the total number of possible matches is?
I've tried to count it in an easy way but I always end up getting confused, can someone help?
Best Answer
A match must be played among four different players.
There are $\binom 6 4 = 15$ different ways to choose four different players.
With any chosen four players, there are exactly three matches to be played (just pair one of them with any of the other three).
Hence the total number of matches is $15 \times 3 = 45$.