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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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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