1. In a season of football matches, the scoring rules are: 3 points for a win; 1 point for a draw; 0 points for a loss. A team has played 15 games and accumulated 33 points. If the order is not considered, the number of possible win, loss, and draw scenarios for the team is ( ).
(A) 3
(B) 4
(C) 5
(D) 6
- 1.A.

Let the team win $x$ games, draw $y$ games, and lose $z$ games. Then $x$, $y$, and $z$ are non-negative integers, and satisfy
$$
\left\{\begin{array}{l}
x+y+z=15, \\
3 x+y=33 .
\end{array}\right.
$$

From equation (2), we get $y=3(11-x)$, substituting into equation (1) yields
$$
z=2(x-9) \text {. }
$$

Also, $0 \leqslant y \leqslant 15,0 \leqslant z \leqslant 15$, thus
$$
0 \leqslant 11-x \leqslant 5,0 \leqslant x-9 \leqslant 7.5 \text {. }
$$

Since $x$ is a non-negative integer, therefore, $9 \leqslant x \leqslant 11$, which means $x$ can only take the values $9$, $10$, and $11$. Correspondingly, the values of $y$ can only be 6, 3, and 0, and the values of $z$ can only be $0$, $2$, and $4$. Therefore, the system of equations (1) and (2) has only three different non-negative integer solutions.