Four teams $A$, $B$, $C$ and $D$ play a football tournament in which each team plays exactly two times against each of the remaining three teams (there are $12$ matches). In each matchif it's a tie each team gets $1$ point and if it isn't a tie then the winner gets $3$ points and the loser gets $0$ points. At the end of the tournament the teams $A$, $B$ and $C$ have $8$ points each. Determine all possible points of team $D$.
Problem
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Tags: combinatorics
WallyWalrus
07.12.2022 21:30
Associate to a team $X\in\{A,B,C,D\}$ the triplet $(W_X;T_X;L_X)=(wins;ties;losses)$. The relations between the results of the teams are: $W_X+T_X+L_X=6\Longrightarrow (W_A+W_B+W_C+W_D)+(T_A+T_B+T_C+T_D)+(L_A+L_B+L_C+L_D)=24 $; $W_A+W_B+W_C+W_D=L_A+L_B+L_C+L_D $. The possible triplets $(W_X;T_X;L_X)$ to obtain $8$ points are $(2;2;2);\;(1;5;0)$. Hence, there are $4$ possible combinations to obtain $8$ points for each of the teams $A,B,C$: $\textbf{Case 1:}\quad(2;2;2);\;(2;2;2);\;(2;2;2)$. Result for the team $D$ the possibilities: $(W_D;T_D;L_D)\in\{(3;0;3);(2;2;2);(1;4;1);(0;6;0)\}$, corresponding the number of points $P_D\in\{9;8;7;6\}$. $\textbf{Case 2:}\quad(2;2;2);\;(2;2;2);\;(1;5;0)$. Result for the team $D$ the possibilities: $(W_D;T_D;L_D)\in\{(2;1;3);(1;3;2);(0;5;1)\}$, corresponding the number of points $P_D\in\{7;6;5\}$. $\textbf{Case 3:}\quad(2;2;2);\;(1;5;0);\;(1;5;0)$. Result for the team $D$ the possibilities: $(W_D;T_D;L_D)\in\{(2;0;4);(1;2;3);(0;4;2)\}$, corresponding the number of points $P_D\in\{6;5;4\}$. $\textbf{Case 4:}\quad(1;5;0);\;(1;5;0);\;(1;5;0)$. Results for the team $D$ the possibility: $(W_D;T_D;L_D)=(0;3;3)$, corresponding the number of points $P_D=3$. Hence, the team $D$ can obtain the number of points $P_D\in\{3,4,5,6,7,8,9\}$.