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Math Help - N dice are thrown

  1. #1
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    N dice are thrown

    I am asked N>=3 sets of dice are thrown. How many way can exactly 3 sixes turn up.

    I did the following

    for a single set of dice....

    dice1+dice2=6 how many solutions?

    (5 choose 1)/2ceiling function??? which is 3 ways

    so for 3 dice to roll 6 there are...

    3x3x3

    then the other n-3 dice can roll anything but a six.

    so there are 18 unique ways to roll a set of dice

    18-3=15 -the total ways to roll a dice with no 6.

    so I get...

    (3)(3)(3)(n-3)(15) ways.

    Now my big problem is how do I express this answer in terms of binomial co-efficient?
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    Re: N dice are thrown

    Quote Originally Posted by ehpoc View Post
    I am asked N>=3 sets of dice are thrown. How many way can exactly 3 sixes turn up.
    This is not a very clear description.
    If N=5, are there 10 dice rolled (5 pair) or are there 5 single dice rolled?
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    Re: N dice are thrown

    LOL I even did a more clear description than the assignment did. Ignorant professor.

    the EXACT question asked

    N ≥ 3 dice are thrown. Express your answers to (a) to (d) in terms of binomial coefficients.
    I inferred that, that meant n sets f dice 2n die in total. Since this is kind of how we have been doing this in class.

    But now I am not sure at all. I am waiting for an email from the prof. Either way the most confusing part is how do I express my answer as a binomial coefficient?
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    Re: N dice are thrown

    If it says "N ≥ 3 dice are thrown. Express your answers to (a) to (d) in terms of binomial coefficients" it means exactly that.
    N dice are thrown. If N=5, then toss 5 dice.
    Now list questions a) - d).
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    Re: N dice are thrown

    1. In how many ways can exactly 3 sixes turn up?
    2. In how many ways can at least 3 sixes turn up?
    3. In how many ways can exactly two 3's and one 4 turn up?
    4. How many combinations of the dice are possible, in total?


    My biggest confusion is mainly in the part of expressing my answer as a niary coefficient
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    Re: N dice are thrown

    Quote Originally Posted by ehpoc View Post
    1. In how many ways can exactly 3 sixes turn up?
    2. In how many ways can at least 3 sixes turn up?
    3. In how many ways can exactly two 3's and one 4 turn up?
    4. How many combinations of the dice are possible, in total?
    This is a mixed set of problems.
    1) \binom{N}{3}\cdot 5^{N-3}

    3) \binom{N}{2}\cdot(N-2)\cdot 4^{N-3}

    Now you tell us about the other two.
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    Re: N dice are thrown

    Question 2

    (N choose 3)*6^n-3?

    4.n!?
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    Re: N dice are thrown

    Hello, ehpoc!

    Do you know what binary coefficients are?


    n \ge 3 dice are thrown.
    Express your answers for (a) to (d) in terms of binomial coefficients.

    (d) How many combinations of dice dice are possible in total?
    There are: . 6^n possible outcomes.



    (a) In how many ways can exactly 3 sixes turn up?
    We want 3 Sixes and n-3 Others.

    There are: . {n\choose3}(1)^3(5)^{n-3} ways.




    (b) In how many ways can at least 3 sixes turn up?

    The opposite of "at at least 3 sixes" is: .0 sixes or 1 six or 2 sixes.

    0 sixes and n Others: . {n\choose0}(1^0)(5^n) \:=\:5^n ways.

    1 six and n-1 Others: . {n\choose1}(1^1)(5^{n-1}) \:=\:n\!\cdot\!5^{n-1} ways.

    2 sixes and n-2 Others: . {n\choose2}(1^2)(5^{n-2}) \:=\:\frac{n(n-1)}{2}\!\cdot\!5^{n-2} ways.

    Less than 3 sixes: . 5^n + n\!\cdot\!5^{n-1} + \frac{n(n-1)}{2}\!\cdot\!5^{n-2} \:=\:\tfrac{1}{2}\,5^{n-2}(n^2 + 9n + 50) ways.

    At least 3 sixes: . 6^n - \left[\tfrac{1}{2}\,5^{n-2}(n^2+9n + 50)\right] ways.




    (c) In how many ways can exactly two 3's and one 4 turn up?

    We want two 3's, one 4, and n-3 Others.

    There are: . {n\choose2,1,n\!-\!3}(1^2)(1^1)(5^{n-3}) ways.

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    Re: N dice are thrown

    Quote Originally Posted by Soroban View Post
    [size=3]There are: . {n\choose2,1,n\!-\!3}(1^2)(1^1)(5^{n-3}) ways.
    Here are a couple of comments (questions?)
    Did you define {n\choose2,1,n\!-\!3}~?

    Is it not {n\choose2,1,n\!-\!3}(1^2)(1^1)({\color{blue}4^{n-3})}~?
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    Re: N dice are thrown

    Ummm I think that answer for B is a little crazy.

    (Nc3) gets our three sixes

    then the other dice can be anything

    6^n-3

    Does this not make sense?

    Also.

    Plato, for you answer for c.

    you have N choose 2

    Is it not N choose 3 still? we still need to choose 3 dice. Two 3's and a 4?

    (nC3)(4)^n-2
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    Re: N dice are thrown

    Quote Originally Posted by ehpoc View Post
    Plato, for you answer for c.
    you have N choose 2
    Is it not N choose 3 still? we still need to choose 3 dice. Two 3's and a 4?
    (nC3)(4)^n-2
    No indeed.
    Choose two places for the two 3's.
    Form the N-2 places left, choose one for the 4.
    In the other N-3 places put any of the four, {1,2,5,6}, values.

    ANSWER: \binom{N}{2}\binom{N-2}{1}(4^{N-3})
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