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?
My attempt:
We consider a team say, $\{P_1, P_2\}.$ Now, we count the number of teams it has to play with. We denote it, by the blank pair $(\__{\text{(1st blank)}},\__{\text{(2nd blank)}}).$ We can place $4$ remaining persons in the 1st blank and $3$ persons in the 2nd. This gives us, $12$ possible arrangements, but then each team is counted twice. So, the number of each teams that $\{P_1, P_2\},$ has to play with is $6.$ Now, this process should be repeated for $\{P_1, P_3\},\{P_1, P_4\},\{P_1, P_5\},\{P_1, P_6\},$ i.e $4$ times. So, the total number of possible matches is $6\times 5=30.$
But the answer given is $45$. I want to know, where's the mistake in my attempt ?
Thank you.
You have missed the $12$ matches that involve $P_2$ but not $P_1$ and the $3$ matches that involve $P_3$ but not $P_1$ or $P_2$.
Another solution is to pick $4$ players from $6$ ($\binom64$) and then arrange the $3$ matches for each.