Question

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

• A. $\sum w_i=\sum l_i=45$
• B. $w_i+l_i=9$
• C. $\sum w_i^2=81+\sum l_i^2$
• D. $\sum w_i^2=\sum l_i^2$

Answer 1

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

The correct options are
A $\sum w_i=\sum l_i=45$
B $w_i+l_i=9$
D $\sum w_i^2=\sum l_i^2$

Each player will play $10-1$ $=9$ games.
Now, there will be total of ${}^{10}{C}_2$$=45$ games.
Each player plays 9 games without any draws. Hence,
$w_i+l_i=9$
The total number of wins $=$ total number of losses. Hence,
$\sum w_i=\sum l_i=45$
$w_i^2=(9-l_i{)}^2$
$=81-18l_i+l_i^2$
$\sum w_i^2=10\left(81\right)-18\left(10\right)\sum l_i+\sum l_i^2$
$=810-18\left(45\right)+l_i^2$
$=\sum l_i^2$
Hence, options A, B and D are correct