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mathmath333

  • one year ago

A mixed double tennis game is to be played between two teams (each team consists of one male and one female ).There are four married couples . No team is to consist of a husband and {his wife. What is the maximum number of games that can be played.

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{A mixed double tennis game is to be played between two teams }\hspace{.33em}\\~\\ & \normalsize \text{(each team consists of one male and one female ).There are }\hspace{.33em}\\~\\ & \normalsize \text{four married couples . No team is to consist of a husband and }\hspace{.33em}\\~\\ & \normalsize \text{his wife. What is the maximum number of games that can be played. }\hspace{.33em}\\~\\ & a.)\ 12 \hspace{.33em}\\~\\ & b.)\ 21 \hspace{.33em}\\~\\ & c.)\ 36 \hspace{.33em}\\~\\ & d.)\ 42 \hspace{.33em}\\~\\ \end{align}}\)

  2. mathstudent55
    • one year ago
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    Husband is A, B, C, D Wife is 1, 2, 3, 4 Married couples are: A1, B2, C3, D4 For each line below, the pair in the first column can play a match with each pair following it. For example, for the first line below, there can be matches: A2 vs B1, A2 vs B3, A2 vs B4, etc. A2 B1 B3 B4 C1 C4 D1 D3 A3 B1 B4 C1 C2 C4 D1 D2 A4 B1 B3 C1 C2 D1 D2 D3 It looks like 21 different matches.

  3. mathmath333
    • one year ago
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    answer is given is 42

  4. mathmath333
    • one year ago
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    i think there are 12 pairs that can play matches with each other

  5. mathmath333
    • one year ago
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    is he still typing

  6. mathstudent55
    • one year ago
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    Sorry. I though all other combinations were repeats. Let me do this again. Here are all possible matches, but some are repeats. A2 B1 B3 B4 C1 C4 D1 D3 A3 B1 B4 C1 C2 C4 D1 D2 A4 B1 B3 C1 C2 D1 D2 D3 B1 A2 A3 A4 C2 C4 D2 D3 B3 A2 A4 C1 C2 C4 D1 D2 B4 A2 A3 C1 C2 D1 D2 D3 C1 A2 A3 A4 B3 B4 D2 D3 C2 A3 A4 B1 B3 B4 D1 D3 C4 A2 A3 B1 B3 D1 D2 D3 D1 A2 A3 A4 B3 B4 C2 C4 D2 A3 A4 B1 B3 B4 C1 C4 D3 A2 A4 B1 B4 C1 C2 C4

  7. mathstudent55
    • one year ago
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    Now we eliminate repeats. A2 B1 B3 B4 C1 C4 D1 D3 A3 B1 B4 C1 C2 C4 D1 D2 A4 B1 B3 C1 C2 D1 D2 D3 B1 C2 C4 D2 D3 B3 C1 C2 C4 D1 D2 B4 C1 C2 D1 D2 D3 C1 D2 D3 C2 D1 D3 C4 D1 D2 D3 Now the total is 42.

  8. welshfella
    • one year ago
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    thats more than 42

  9. welshfella
    • one year ago
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    sorry - that is 42

  10. thomas5267
    • one year ago
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    4 man and 4 women can form 16 teams. 4 teams are forbidden since there are 4 married couples. There are 12C2=66 unique ways of choosing 2 elements from a set with 12 elements. Therefore 12 teams can have 12C2=66 ways of matching and game.

  11. thomas5267
    • one year ago
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    The set of available matches are: {{1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7}, {1, 8}, {1, 9}, {1,10}, {1, 11}, {1, 12}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {2, 7}, {2, 8}, {2, 9}, {2, 10}, {2, 11}, {2, 12}, {3, 4}, {3, 5}, {3, 6}, {3, 7}, {3, 8}, {3, 9}, {3, 10}, {3, 11}, {3, 12}, {4, 5}, {4, 6}, {4, 7}, {4, 8}, {4, 9}, {4, 10}, {4, 11}, {4, 12}, {5, 6}, {5, 7}, {5, 8}, {5, 9}, {5, 10}, {5, 11}, {5, 12}, {6, 7}, {6, 8}, {6, 9}, {6, 10}, {6, 11}, {6, 12}, {7, 8}, {7, 9}, {7, 10}, {7, 11}, {7, 12}, {8, 9}, {8, 10}, {8, 11}, {8, 12}, {9, 10}, {9, 11}, {9, 12}, {10, 11}, {10, 12}, {11, 12}}

  12. thomas5267
    • one year ago
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    The size of the set is 66.

  13. mathmath333
    • one year ago
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    mathstudent's answer is correct, but i m looking for a short answer rather than listing all as mathstudent did.

  14. anonymous
    • one year ago
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    @thomas5267 only problem is that A2 cannot play A3. In your scheme, these are indeed different teams but they cannot play each other. For a given team, it has only 6 other teams against which it can compete. then you have to count repeats as well.

  15. anonymous
    • one year ago
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    oops, 7 other teams against which it can compete

  16. thomas5267
    • one year ago
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    Oh crap!

  17. mathstudent55
    • one year ago
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    @thomas5267 The problem states that a man and his wife cannot form a team, so you can't form 16 teams.

  18. thomas5267
    • one year ago
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    Computer says there are 42 combinations. I calculated it using brute force.

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