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Is there a way to find the combinations without listing the outcomes

  1. Nov 18, 2009 #1
    Here is a list of sums rolled by 4 six-sided dice. The number of combinations for each sum is given below each sum.


    ...........sums: 4 5 6 7 8 9 10 11 12 13 14
    combinations: 1 4 10 20 35 56 80 104 125 140 146

    ...........sums: 15 16 17 18 19 20 21 22 23 24
    combinations: 140 125 104 80 56 35 20 10 4 1


    I notice that the combination 146 is symmetric about the sum of 14.
    I mean each combination for each sum less than 14 is the same as each combination for each sum greater than 14.

    Is it possible to find a formula or use Pascal s triangle or anything to find the combination for any sum here WITHOUT listing any outcomes.

    How can it be done Thanks.
     
  2. jcsd
  3. Nov 18, 2009 #2
    With a recursion relation and a generating function I figured out, that these numbers are always the coefficients of the powers in
    [tex]
    f(m)=(z+z^2+z^3+z^4+z^5+z^6)^m=\left(z\frac{z^6-1}{z-1}\right)^m=z^m(1+z+z^2)^m(1+z^3)^m=z^m(1+z^2+z^4)^m(1+z)^m
    [/tex]
    with m as the number of dice (you can check with Wolfram Alpha). Now it depends on how easy it is to extract one of these coefficients...
    That's basically a very advanced Pascal triangle :smile:

    Edit:
    I tried to find a closed form answer with the above method. I don't have Mathematica to check, but the actual answer can't be much off
    [tex]
    N(x)=\sum_{a=0,b=0}^m (-1)^{a+b}3^{m-b}\binom{m}{a}\binom{m}{b}\binom{2a+2b+m}{x+a+b-3m}
    [/tex]
    for the number of combinations to get sum x with m dice. I guess the binomial with negative numbers should be equal to zero.
     
    Last edited: Nov 18, 2009
  4. Nov 19, 2009 #3
    The lazy person's solution: type

    expand (x + x^2 + x^3 + x^4 + x^5 + x^6)^4

    into Wolfram Alpha and hit Enter.
     
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