Combinations for Dice

[Soulkeeper]

(N+5) (N+4) (N+3) (N+2) (N+1) / 120

(6+5) (6+4) (6+3) (6+2) (6+1) / 120
11 * 10 * 9 * 8 * 7 / 120 = 462
10 * 9 * 8 * 7 * 6 / 120 = 252
9 * 8 * 7 * 6 * 5 / 120 = 126
8 * 7 * 6 * 5 * 4 / 120 = 56
7 * 6 * 5 * 4 * 3 / 120 = 21
6 * 5 * 4 * 3 * 2 / 120 = 6
923

5! = 5 * 4 * 3 * 2 * 1 = 120
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720

Am I correct in using 5! for the calculation of combinations for a variable number of 6-sided dice (N=dice) ?

 

[CycloWizard]

For dice with different numbers of sides, it'is just d1*d2*d3*...*dN. The specific case where all dice have the same number of sides is then d^N, where d is the number of sides on each die. I think...

 

[Soulkeeper]

For dice with different numbers of sides, it'is just d1*d2*d3*...*dN. The specific case where all dice have the same number of sides is then d^N, where d is the number of sides on each die. I think...

d^N would be permutations
I need combinations where order is not important

 

[wuliheron]

That gives you all the possible combinations, but you need to divide the result by 2 to get distinct permutations.

 

[Soulkeeper]

That gives you all the possible combinations, but you need to divide the result by 2 to get distinct permutations.

That would be incorrect also, take for example one 6-sided dice (6^1)/2 = 3

unless you mean something else ...

 

[TecHNooB]

For dice with different numbers of sides, it'is just d1*d2*d3*...*dN. The specific case where all dice have the same number of sides is then d^N, where d is the number of sides on each die. I think...

this is right.

 

[Soulkeeper]

this is right.

permutations ie:
6^1
6^2
6^3
...

I'm dealing with the "specific case" where all dice have the same number of sides and don't want the permutations.

 

[Soulkeeper]

for 6 dice:
(N+5) (N+4) (N+3) (N+2) (N+1) / 120
(6+5) (6+4) (6+3) (6+2) (6+1) / 120
11 * 10 * 9 * 8 * 7 / 120 = 462 [combinations]

I just need someone to verify or refute this.

 

[Soulkeeper]

I think I just verified my equation

http://pastebin.com/AqTiC4yC
that's for 2 dice, already knowing the equation works with 1 dice.