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If Janet and Sylvia played the same 4 people in a game, Marilyn, Amy, Emmy, and Pam,...

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moocow554 | Student, Undergraduate | Valedictorian

Posted February 22, 2013 at 11:28 PM via web

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If Janet and Sylvia played the same 4 people in a game, Marilyn, Amy, Emmy, and Pam, and Janet's scores were, 467 against Marilyn, 448 against Amy, 431 against Emmy and 454 against Pam,  and Sylvia's Scores were 379 against Marilyn, 430 against Amy, 340 against Emmy, and 365 against Pam,  what would be the best combination of points using 2 of Janet's scores and 2 of Sylvia's scores? Make sure there is a score from all 4 opponents, Marilyn, Amy, Emmy, and Pam. 

 
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pramodpandey | College Teacher | Valedictorian

Posted March 24, 2013 at 10:29 AM (Answer #2)

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If Janet and Sylvia played the same 4 people in a game, Marilyn, Amy, Emmy, and Pam

           M          A          E           P

         Marilyn    Amy     Emmy     Pam

Janet   467      448        431       452

Sylvia   379      430       340       365 

Possible combination of Janet C(4,2)= 6

             MA     ME     MP     AE     AP       EP

Janet     915   898    921    879   902     885

Sylvia     809   719    744    770   795    705

Under given condition we can select

MAEP   MEAP  MPAE   AEMP   APME  EPMA

1620    1693  1691    1623   1621   1694

Thus best combination is  Janet EP ( Emmy and Pam) and Sylvia

(Marilyn    Amy ) . Best score =1694

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