Question

Ten persons numbered 1, 2, 3…………10 plays a chess tournament, each player playing against every other player in 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 ${l_1},{l_2},{l_{3,}}...............{l_{10}}$ are the number of games lost by the players 1, 2, 3,………………… 10 respectively then
$(a){\text{ }}\sum {{w_i} = \sum {{l_i} = 45} } \\\ (b){\text{ }}{{\text{w}}_i} + {l_i} = 9 \\\ (c){\text{ }}\sum {{w_i}^2 = 81 + \sum {{l_i}^2} } \\\ (d){\text{ }}\sum {{w_i}^2 = \sum {{l_i}^2} } \\\$

Answer 1

Hint – In this question first compute the total number of games played by each of these 10 players and it will be (10-1=9). Now each player plays against one another and a match is held between two so find the total number of games. Since no match can end in a draw, the sum of winning and losing for a particular ${i^{th}}$ player will be equal to the total game that can be played by a single player. This concept will help getting the right option.

Complete step-by-step answer:

There are a total of 10 players.
So each player will play (10 – 1) = 9 games.
Now there are 10 players and a game of chess is played between two candidates and it is given that each player will play against every other player so the total number of games = ${}^{10}{C_2} = \dfrac{{10!}}{{2!\left( {10 - 2} \right)!}} = \dfrac{{10 \times 9 \times 8!}}{{2 \times 1 \times 8!}} = 45$
Now $w_1, w_2, ......., w_{10}$ and $l_1, l_2, ....., l_{10}$ are the number of games won and lost by players 1, 2, ...., 10 respectively.
Now each player plays 9 games without any draws.
Hence,
${w_i} + {l_i} = 9$....................... (1) (i.e. it’s the sum of winning and losing a game for a particular ith player is equal to 9).
Now the total number of wins = total number of losses.
$\Rightarrow \sum {{w_i} = \sum {{l_i}} }$....................... (2)
Now apply summation on both sides in equation (1) we have,
$\Rightarrow \sum {\left( {{w_i} + {l_i}} \right)} = \sum 9$
$\Rightarrow \sum {{w_i} + \sum {{l_i} = 9\left( {10} \right) = 90} }$ , as there are total 10 players therefore, $\sum 9 = 9\left( {10} \right) = 90$
Now from equation (2) we have,
$\Rightarrow \sum {{w_i} + \sum {{w_i} = 90} }$
$\Rightarrow \sum {{w_i}} = \dfrac{{90}}{2} = 45$
Hence,
$\sum {{w_i} = \sum {{l_i} = 45} }$......................... (3)
Now from equation (1) we have,
$\Rightarrow {w_i} = 9 - {l_i}$
Now squaring on both sides we have,
$\Rightarrow {\left( {{w_i}} \right)^2} = {\left( {9 - {l_i}} \right)^2}$
Now expand the square according to property ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ so we have,
$\Rightarrow {\left( {{w_i}} \right)^2} = {9^2} + {\left( {{l_i}} \right)^2} - 18{l_i}$
Now take summation on both sides we have,
$\Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = \sum {\left( {{9^2} + {{\left( {{l_i}} \right)}^2} - 18{l_i}} \right)}$
$\Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = \sum {81} - 18\sum {\left( {{l_i}} \right)} + \sum {{{\left( {{l_i}} \right)}^2}}$
Now there are total 10 players
Therefore $\sum {81} = 81\left( {10} \right) = 810$
$\Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = 810 - 18\sum {\left( {{l_i}} \right)} + \sum {{{\left( {{l_i}} \right)}^2}}$
Now from equation (3) we have,
$\Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = 810 - 18\left( {45} \right) + \sum {{{\left( {{l_i}} \right)}^2}}$
$\Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = 810 - 810 + \sum {{{\left( {{l_i}} \right)}^2}}$
$\Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = \sum {{{\left( {{l_i}} \right)}^2}}$
So this is the required answer.
Hence option (A), (B) and (D) are correct.

Note – Now as the matches proceed the total number of winning has to be equal to the total number of losing , as when a match is played between two players there can only be two outcomes either person one wins and person 2 loses or person 1 loses and person 2 wins. Since the drawn cases need not to be taken into consideration thus under this circumstance only this equation holds true, and it’s the trick point of this problem.