Question

Ten persons numbered $1,2,\cdots ,10$ play a chess tournament, each player playing against every other exactly one game. It is known that no game ends in a draw. Let $w_1,w_2,....,w_{10}$ are the number of games won by players $1,2,3,\cdots ,10$, respectively, and $l_1,l_2,\cdots ,l_{10}$ are the number of games lost by the players $1,2,\cdots ,10$, respectively, then,

• A. $\sum _{i=1}^{10}{w}_i+\sum _{i=1}^{10}{l}_i=90$
• B. $w_i+l_i=9$, where $i\in \{1,2,\cdots ,10\}$
• C. $\sum _{i=1}^{10}{w}_i^2=81+\sum _{i=1}^{10}{l}_i^2$
• D. $\sum _{i=1}^{10}{w}_i^2=\sum _{i=1}^{10}{l}_i^2$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\sum _{i=1}^{10}{w}_i+\sum _{i=1}^{10}{l}_i=90$
B $w_i+l_i=9$, where $i\in \{1,2,\cdots ,10\}$
D $\sum _{i=1}^{10}{w}_i^2=\sum _{i=1}^{10}{l}_i^2$

Clearly, each player will play 9 games. And total number of games is ${}^{10}{C}_2=45.$
Clearly,
$w_i+l_i=9\text{and}\sum _{i=1}^{10}{w}_i=\sum _{i=1}^{10}{l}_i=45$
$\Rightarrow w_i=9-l_i\Rightarrow w_i^2=81+l_i^2-{181}_i$
$\Rightarrow \sum _{i=1}^{10}{w}_i^2=81\times 10+\sum _{i=1}^{10}{l}_i^2-18\sum _{i=1}^{10}{l}_i=810+\sum _{i=1}^{10}{l}_i^2-18\times 45=\sum _{i=1}^{10}{l}_i^2$