You are looking for combinations. Just with additional constraints. Look at a simpler example - 2 dice. Your rules says that 1-3 and 3-1 are the same. So, we take the first number in the first die and look at the possibilities in the second - 6. Then we take the second number in the first die, leaving us 5 possibilities in the second die. (The sixth, 2-1, is disallowed). Following the pattern leaves us 6+5+4+3+2+1 = 21 possibilities.

Extending to three dice, we have the following - set to 1-1 and we have 6 possibilities. 1-2 and we have 5, etc. So, with the first die as 1, we have the above 21 possibilities. Setting the first die to 2 and we have 5 possibilities for the second and 4 for the third - leaving us 15 total possibilities. So, the total ends up being 21 + 15 + 10 + 6 + 3 + 1 = 56, which is borne out by your code.

You can extend the pattern upwards. The actual formula involves a bunch of Sigmas. 

d = 6, n = 2 ==> S(i=1->d)(i)
d = 6, n = 3 ==> S(i=1->d)(S(j=1->i)(j)

The inductive rule is:
F(2) ==> S(i=1->d)(i)
F(n) ==> S(i=1->d)(F(n-1))

[download]

I'm sure there's a straight formula, but my brain hurts. :-)

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Then there are Damian modules.... *sigh* ... that's not about being less-lazy -- that's about being on some really good drugs -- you know, there is no spoon. - flyingmoose

I shouldn't have to say this, but any code, unless otherwise stated, is untested

In reply to Re: unordered sets of N elements by dragonchild
in thread unordered sets of N elements by japhy

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