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Ten persons numbered 1, ,2 ..,10 play...

Ten persons numbered 1, ,2 ..,10 play a chess tournament, each player 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. `sumw_1=suml_i=45` b. `w_1+1_i=9` c. `sumw l1 2()_=81+suml1 2` d. `sumw l i2()_=suml i2`

A

`sumw_(i)^(2)+81-suml_(i)^(2)`

B

`sumw_(i)^(2)+81=suml_(i)^(2)`

C

`sumw_(i)^(2)=suml_(i)^(2)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
Clearly, each player will play 9 games.
`therefore`Total number of games=`.^(10)C_(2)=45`
clearly, `w_(i)+l_(i)=9 and sumw_(i)=suml_(i)=45`
`impliesw_(i)=9-l_(i)impliesw_(1)^(2)=81-18l_(i)+l_(i)^(2)`
`impliesumw_(i)^(2)=sum81-18suml_(i)+suml_(1)^(2)`
`=81xx10-18xx45+suml_(i)^(2)=suml_(1)^(2)`.
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