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Okay, here is a problem that has been bugging me:

So far, I believe I have found that answering the main question is equivalent to determining the number of (

{{

Basically, starting with each die at one and counting the number of ways I can build up to the total,

I can find a solution where

If someone could help me finish my solution or propose an alternate, easier solution, I would be very grateful.

(Added:)

Basically, if we are using two six-sided dice, then there are six ways to add up to seven:

1+6, 2+5, 3+4, 4+3, 5+2, and 6+1.

And five ways to add up to eight:

2+6, 3+5, 4+4, 5+3, and 6+8.

Et cetera.

I'm not necessarily looking for a general solution, though one would be great. But I would like to at least be able to compute the answer for particular sets of input.Say that you havendice that each havedsides, numbered 1 throughm. How many different dice combinations, out of a total ofnxdpossible, are there where the sum of all of the dice is equal tor, for particularn,d, andr?

So far, I believe I have found that answering the main question is equivalent to determining the number of (

*r*−*n*) -combinations in a multiset with the form{{

*k*_{1}, ... ,*k*_{n}}, {(*k*_{1},*d*−1), ... , (*k*_{n},*d*−1)}}Basically, starting with each die at one and counting the number of ways I can build up to the total,

*r*, without exceeding*d*in any of them.I can find a solution where

*d*= ∞ easily enough, but I am not sure how to eliminate the illegal combinations from that using Inclusion-Exclusion.If someone could help me finish my solution or propose an alternate, easier solution, I would be very grateful.

(Added:)

Basically, if we are using two six-sided dice, then there are six ways to add up to seven:

1+6, 2+5, 3+4, 4+3, 5+2, and 6+1.

And five ways to add up to eight:

2+6, 3+5, 4+4, 5+3, and 6+8.

Et cetera.

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