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Math Help - Posible Dominoes Games

  1. #1
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    Posible Dominoes Games

    Hi and I would really appreciate if somebody help me on this.

    I want to find out how many dominoes games are posible to do with a set of 28 tiles and every game including 4 players in which each takes 7 tiles (4x7=28)

    I was thinking the solution would be (Z):
    C (28,7) x C (21,7) x C (14,7)= 4.72E+14=Z

    I dont think the number is correct as if I pick a game random from Z and I call it R and I call the four players R1, R2, R3 and R4.

    Lets say that our current game R made of R1, R2, R3 and R4 is changed to R4, R2, R1 and R3 (the players simply changed positions in the table, keeping their tiles). For my purposes, this is a different game, even though the players still have the same tiles but the fact that they are in a different positions, it creates a different game as the turns are anti clock.

    I wonder if I have to do a permutation of 4 which is equal to 24.

    Then my number of games could be equal to 24 x 4.72E+14

    is this correct? or are those variations already on my previous solution?

    thanks for the help.
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  2. #2
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    Hello Verdugo

    Welcome to Math Help Forum!
    Quote Originally Posted by Verdugo View Post
    Hi and I would really appreciate if somebody help me on this.

    I want to find out how many dominoes games are posible to do with a set of 28 tiles and every game including 4 players in which each takes 7 tiles (4x7=28)

    I was thinking the solution would be (Z):
    C (28,7) x C (21,7) x C (14,7)= 4.72E+14=Z

    I dont think the number is correct as if I pick a game random from Z and I call it R and I call the four players R1, R2, R3 and R4.

    Lets say that our current game R made of R1, R2, R3 and R4 is changed to R4, R2, R1 and R3 (the players simply changed positions in the table, keeping their tiles). For my purposes, this is a different game, even though the players still have the same tiles but the fact that they are in a different positions, it creates a different game as the turns are anti clock.

    I wonder if I have to do a permutation of 4 which is equal to 24.

    Then my number of games could be equal to 24 x 4.72E+14

    is this correct? or are those variations already on my previous solution?

    thanks for the help.
    You are right. You can divide the 28 dominoes into 4 groups of 7 in \binom{28}{7}\binom{21}{7}\binom{14}{7} ways, and then allocate these groups to the four players in 4! ways. Thus the total is
    4!\binom{28}{7}\binom{21}{7}\binom{14}{7}\approx 1.134\times 10^{16}
    Grandad
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    Quote Originally Posted by Grandad View Post
    Hello Verdugo

    Welcome to Math Help Forum!You are right. You can divide the 28 dominoes into 4 groups of 7 in \binom{28}{7}\binom{21}{7}\binom{14}{7} ways, and then allocate these groups to the four players in 4! ways. Thus the total is
    4!\binom{28}{7}\binom{21}{7}\binom{14}{7}\approx 1.134\times 10^{16}
    Grandad
    Thank you very much for your help.
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    Sorry for activating this tread again but I have people telling me that my calculation is wrong.

    the portion that is aparently wrong is the 4!.

    What I have been told is that rather than multiplying for 4!, I should just multiply by 4.

    Can anyone give me a second opinion on this?

    I'm not sure if anyone here can read spanish but I have an online book in spanish also mentioning that the number to use is 4 rather than 4!.

    the formula is used on page 54-55 of this book preview:
    El arte del dominó: teoría y práctica - Google Books
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    any second opinion?
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  6. #6
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    Verdugo,

    If I understand your problem correctly, then your original answer was correct.

    The first player can choose his dominoes in \binom{28}{7} ways; then the second player can choose his in \binom{21}{7} ways; then the third player can choose his in \binom{14}{7} ways. The fourth player has no choice and get what dominoes are left. So the total number of possibilities is

    \binom{28}{7}\binom{21}{7} \binom{14}{7}.

    Another way to look at the answer is that it is a mutinomial coefficient,
    \binom{28}{7 \; 7 \; 7 \; 7}
    which is really the same result, just written differently.

    I am assuming that the assignment of dominoes to players is significant, i.e. if two players swap their sets of dominoes then the arrangement is always counted as different. If this is not what you meant, then please explain.
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    Quote Originally Posted by awkward View Post
    Verdugo,

    If I understand your problem correctly, then your original answer was correct.

    The first player can choose his dominoes in \binom{28}{7} ways; then the second player can choose his in \binom{21}{7} ways; then the third player can choose his in \binom{14}{7} ways. The fourth player has no choice and get what dominoes are left. So the total number of possibilities is

    \binom{28}{7}\binom{21}{7} \binom{14}{7}.

    Another way to look at the answer is that it is a mutinomial coefficient,
    \binom{28}{7 \; 7 \; 7 \; 7}
    which is really the same result, just written differently.

    I am assuming that the assignment of dominoes to players is significant, i.e. if two players swap their sets of dominoes then the arrangement is always counted as different. If this is not what you meant, then please explain.
    If I understand your answer correctly, I do not need to multiply Z by 4!?

    I will explain a bit further.

    After each player has selected his/her 7 tiles (set). Lets call them S1, S2, S3 and S4.

    These sets can be combined in 4! like: S1-S2-S4-S3, S2-S3-S1-S4, ect. All these combinations (4!) are different games for me.
    Last edited by Verdugo; December 13th 2009 at 04:34 AM.
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  8. #8
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    Hello, Verdugo!

    I want to find out how many dominoes games are posible to do with a set of 28 tiles.
    Every game has 4 players, each taking 7 tiles.

    I was thinking the solution would be:
    . . C(28,7) \times C(21,7) \times C (14,7)\:=\: 4.72 \times 10^{14}
    I believe that this answer is correct.


    Call the players \{A,B,C,D\}

    There are: . C(28,7) ways to give 7 tiles to player A.

    Then there are: . C(21,7) ways to give 7 tiles to player B.

    Then there are: . C(14,7) ways to give 7 tiles to player C.

    Then the remaining 7 tiles are given to player D.


    So you have already included the ordering of the players.



    Another approach: . {28\choose7,7,7,7} .which produces the same answer.

    This is an ordered partition of 28 objects into four equal groups.

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    Quote Originally Posted by Verdugo View Post
    If I understand your answer correctly, I do not need to multiply Z by 4!?

    I will explain a bit further.

    After each player has selected his/her 7 tiles (set). Lets call them S1, S2, S3 and S4.

    These sets can be combined in 4! like: S1-S2-S4-S3, S2-S3-S1-S4, ect. All these combinations (4!) are different games for me.
    You do not need to multiply by 4!.

    But I don't understand what you mean by
    S1-S2-S4-S3, S2-S3-S1-S4, ect. All these combinations (4!) are different games for me.
    If S1, S2, S3, and S4 are sets of dominoes, what is S1-S2-S4-S3?
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    Quote Originally Posted by awkward View Post
    You do not need to multiply by 4!.

    But I don't understand what you mean by
    If S1, S2, S3, and S4 are sets of dominoes, what is S1-S2-S4-S3?
    Sorry for not explaining myself bettter.

    S1-S2-S3-S4. Are the 7 tiles that each player has. Player 1 has seven tiles called "S1", player 2 has seven tiles called "S2" and so on.

    So, there are 24 ways in which this game can be combined.

    For the purpose of this tread, player 1 can trade his tiles with player 2. creating S2-S1-S4-S3 from a game that was originally S1-S2-S3-S4.
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  11. #11
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    well I think I found my answer.

    I decided to test something. I made a sample game with 6 tiles and 3 players in which every player takes 2 tiles
    C (6,2 ) x C(4 , 2) x C (2,2)

    the tiles are:A,B,C,D,E and F

    the main games are:

    1) AB,CD, EF
    2) AB, CF, DF
    3) AC, BD, EF
    4) AC, BE, DF
    5) AD, BC, EF
    6) AD, BF, CE

    and so on...

    Now, I can permutate any of this games. For instance game 1:
    1-2) AB, EF, CD
    1-3) CD, AB, EF
    1-4) CD, EF, AB
    1-5) EF, AB, CD
    1-6) EF, CD, AB
    Last edited by Verdugo; December 13th 2009 at 06:36 AM.
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