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

[davedave]

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.

 

[Gerenuk]

With a recursion relation and a generating function I figured out, that these numbers are always the coefficients of the powers in
$$f(m)=(z+z^2+z^3+z^4+z^5+z^6)^m=\left(z\frac{z^6-1}{z-1}\right)^m=z^m(1+z+z^2)^m(1+z^3)^m=z^m(1+z^2+z^4)^m(1+z)^m$$
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

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
$$N(x)=\sum_{a=0,b=0}^m (-1)^{a+b}3^{m-b}\binom{m}{a}\binom{m}{b}\binom{2a+2b+m}{x+a+b-3m}$$
for the number of combinations to get sum x with m dice. I guess the binomial with negative numbers should be equal to zero.

 

[awkward]

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.