# Multiple Dice Permutations for Different Number Ranges Each Die

Looking how to calculate (1) the number of permutations for different number ranges on each dice; and (2) the number of permutations for number ranges shared by two or more dice.

All dice are fair and 6 sided.

I can do the easy calculations where how many rolls have exactly one 6, or at least one 6, two 6's, or at least two 6's etc.

'm looking for a more general approach where I have m dice, and find permutations where say where:

at least one die is 5+ and at least another is 4+.

in general I would like to be able to find a formula or counting approach what are the number of permutations where x die are greater than 5, when rolling m dice where a constant c can be distributed across the die.

• Any help in setting up the problem would be useful as well. Currently looking at defining it as a set S, with the elements S={1,2,3,4,5,6}. Each dice "roll" corresponds a selection from the set. So thinking the problem could be phrased as how many permutations exist where on three selections (s1,s2,s3) at least one selection is a {4,5,6} and another is {5,6}. – MikeX Nov 6 '17 at 15:34
• A more enumeration approach I am now trying makes use of generating functions. Let each 6-sided die have its own generating function. I assign generating functions as follows. For the die I want a 4,5 or 6 on it is 3x+3y. For the other two die I want at least 5+, so I assign for their generating functions 4x+2y. Then I just sum the coefficients of any term containing y. I've tired this for 2 dice and 3 dice with a simpler case (6+ on at least one die) and it seems to work. Since the math is beyond me, going to try brute force enumeration of all 3 dice permutations and check. – MikeX Nov 8 '17 at 2:51

I found a method to answer this question using generating functions. Represent set each die by the equation $3x+2y+z$ where x represents the numbers 1-3, y represents the numbers 5-6, and z the number 4.
For m dice calculate the polynomial $(3x+2y+z)^m$
for example, with 4 dice the sum of the coefficients of the terms $y^4$ and $y^3z$ will give you the number of combinations you can have 4 dice with a 5 or higher when you can add +1 to one of the rolls.
The sum of the coefficients of the terms $xy^3+xy^2z+y^2z^2$ will provide the number of combinations with 3 dice 5 or higher where you can add +1 to one die.