Graph Theory: A Tournament Question First of all, this is a homework question:
In a tournament which 18 teams participate, a team being matched with another in a
round don't match again in the following (later) rounds. After 8 rounds prove that
there are 3 teams not being matched with each other.

I know there is a so-called solution here pigeonhole principle related problem but this does not explain "prove ... 3 teams not being matched with each other." part which is the core of the question. 
By the way, this part asks to prove that 3 teams, say A,B,C, will play each other after round 8. (ie. show that there are A-B,A-C and B-C matches after round 8.)
Please, any help-hint to solve second part will be appreciated!

Suppose that there are $9$ male teams and $9$ female teams in the tournament, and in the first $8$ rounds it just happens that all of the male teams play one another and all of the female teams play one another. In other words, it’s as if the first $8$ rounds were two round-robin tournaments running separately, one for the men and one for the women. That accounts for $2\binom92=72$ matches. Now pick any $3$ teams: at least two of them must be of the same sex, and those two have already played each other. Thus, it’s impossible in this case to find $3$ teams, no two of which have already played each other.
Added: However, as mualloc has pointed out, this scheme is actually not possible: $9$ is odd, so it would leave one male team and one female team sitting out each round. 
Suppose that in every set of $3$ teams, two of the teams played each other in the first $8$ rounds. Let $v$ be one of the teams, and let $u_1,\ldots,u_9$ be the $9$ teams that $v$ has not played after the first $8$ rounds. Then each pair of teams in the set $\{u_1,\ldots,u_9\}$ must have played each other: if $u_k$ and $u_\ell$ have not played each other, then no two of $v,u_k$, and $u_\ell$ have played each other. But then the games that the $u_k$ played against one another account for all of their games, so the other $9$ teams must have played all of their games against one another as well, and this is the situation which we just saw is impossible. Thus, there must be a set of $3$ teams, no two of which have played one another.
