7) At the end of a football league with 20 teams (where each team plays against every other team exactly two matches), the Mathematicians won 19 matches, drew 12, and lost 7. The coach observes that $19=12+7$. For a generic team in the league, we denote with $(v, n, p)$ the triplet formed by the number of wins, draws, and losses respectively, achieved in the league. For how many distinct triplets can it happen that $v=n+p$? (Note: The triplets are ordered, so, for example, $(19,12,7)$ and $(19,7,12)$ are to be considered distinct). (A) 19, (B) 20, (C) 38, (D) 40, (E) none of the above.
7. The answer is (B). Each team plays 38 matches, so $v+n+p=38$. If $v=n+p$ then $v=19$ and also $n+p=19$. The triples for which $v=n+p$ are then:

$$
(19,19,0),(19,18,1),(19,17,2), \ldots,(19,1,18),(19,0,19)
$$

In total, there are 20 distinct ordered triples.