CMI 2024 | Data Science | Part A | Question: 4

Question

$14$ teams participate in a volleyball tournament. Each team plays the other exactly once. There are no draws. Assume the teams are labeled by $j, 1 \leq j \leq 14$. Let $x_{j}$ denote the number of games team $j$ wins and $y_{j}$ the number of games that team $j$ lost. Pick the correct alternative(s):

• A. $\sum_{j} x_{j}^{2}=\sum_{j} y_{j}^{2}$.
• B. $\sum_{j} x_{j}^{2}>\sum_{j} y_{j}^{2}$.
• C. $\sum_{j} x_{j}=\sum_{j} y_{j}$.
• D. $\sum_{j}\left|x_{j}\right|=\sum_{j}\left|y_{j}\right|$.

   

Answer 1

In this tournament with 14 teams, each team plays every other team once.
So, each team plays 13 matches, and for each team j:
$ x_j + y_j = 13 $, where $x_j$ = wins, $y_j$ = losses.

Total number of games will e:
$ \binom{14}{2} = 91 $

Each game gives one win and one loss  
So, total wins = total losses = 91

Hence  $ \sum_{j} x_j = \sum_{j} y_j = 91 $

 

also $\ y_j = 13 - x_j \Rightarrow y_j^2 = (13 - x_j)^2 $

So,
$ \sum_{j} y_j^2 = \sum (13 - x_j)^2 = \sum_{j} (169 - 26x_j + x_j^2) $
$ = 14 \cdot 169 - 26 \sum_{j} x_j + \sum_{j} x_j^2 $
$ = 2366 - 26 \cdot 91 + \sum_{j} x_j^2 = 2366 - 2366 + \sum_{j} x_j^2 = \sum_{j} x_j^2 $

$ \sum_{j} x_j^2 = \sum_{j} y_j^2 $

 

Since $x_j, y_j \ge 0$:

$ \sum_{j} |x_j| = \sum_{j} x_j = 91 $  
$ \sum_{j} |y_j| = \sum_{j} y_j = 91 $
 

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$ \boxed{\text{A, C, D}} $