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Math Help - 2 question in permutations and Combination

  1. #1
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    2 question in permutations and Combination

    1 ) there are 10 teams in a league .if each team is to play evcry other team exactly once .
    A ) how many games must be scheduled ?
    B ) How many games must be on each team ?

    A ) 10C9 = 10
    B ) 10C1 = 10

    2 )_A catering service offers 8 appetize, 10 main courses and 7 desserts . A Chairperson is to select
    3 appertizs , 4 main courses and 2 desserts for A dinner . How many way can this be done ?

    Here in this question we use C or P and why ?
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  2. #2
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    Quote Originally Posted by r-soy View Post
    1 ) there are 10 teams in a league .if each team is to play evcry other team exactly once .
    A ) how many games must be scheduled ?
    B ) How many games must be on each team ?

    A ) 10C9 = 10
    B ) 10C1 = 10

    2 )_A catering service offers 8 appetize, 10 main courses and 7 desserts . A Chairperson is to select
    3 appertizs , 4 main courses and 2 desserts for A dinner . How many way can this be done ?

    Here in this question we use C or P and why ?
    hi

    (2) That would be combination because take dessert for example , it can be ice cream , pudding , cakes , ... the order is not important here , it could be ice cream , cakes , pudding , OR pudding , ice , cakes , its the same dessert we are talking about .

    so 8C3 x 10C4 x 7C2
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  3. #3
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    Quote Originally Posted by r-soy View Post
    1 ) there are 10 teams in a league .if each team is to play evcry other team exactly once .
    A ) how many games must be scheduled ?
    B ) How many games must be on each team ?

    A ) 10C9 = 10
    B ) 10C1 = 10

    [/COLOR]
    (A)

    TeamA plays the other 9 teams

    TeamB has played TeamA and must play the other 8 teams

    TeamC has played both TeamA and TeamB, so has the other 7 teams to play.

    This is 9+8+7+6+5+4+3+2+1=45 games.

    Or...each time, 5 games can be played (say on a Saturday).
    Each team must play 9 games.

    That's 9(5)=45 games
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  4. #4
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    But in like this queation we solve by rule

    i mean we use C
    for example 10 C 9

    i want expline the solution like this way
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  5. #5
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    Quote Originally Posted by r-soy View Post
    But in like this queation we solve by rule

    i mean we use C
    for example 10 C 9

    i want expline the solution like this way
    Ok,

    in that case, in terms of "selections",
    we need to calculate the number of ways of pairing off 2 teams from 10.

    That will give us the number of games playable in a league of 10 teams.

    \binom{10}{2}=10c_2=\frac{10!}{(10-2)!2!}=\frac{10!}{8!2!}=\frac{10(9)}{2}=5(9)=45
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  6. #6
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    Quote Originally Posted by r-soy View Post
    1 ) there are 10 teams in a league .if each team is to play evcry other team exactly once .

    B ) How many games must be on each team ?


    B ) 10C1 = 10
    Hi r-soy,

    could you be more specific when you say "on each team" ?

    Are you asking

    "Write in terms of factorial notation, the number of games played by each team" ?

    If so, this is 9c_9,\ or\ \binom{9}{9} as each team plays the 9 others (they cannot play against themselves).
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