Very simple (?) dice question

Question

We were just discussing with colleagues the number of combinations you could get with two "normal", $6$-sided dice. Almost all of my colleagues were saying $36$ ($6^2$), which I agree with as such, but you will get almost half of the possible combinations counted twice. If I count, the number of different combinations with $2$ dice is $21.$ I'm not able though to get to the formula that would let me calculate this for $3,4,5$ or more dice.

I remember from my old mathematics courses that there are several different formulas you can "pick" depending on repetition, order importance, etc... What would be the appropriate formula taking into account $n$ as the number of dice and $s$ the number of sides ?

Answer 1

It agrees with the number of sums $a_1+\cdots+a_s=n$ where the $a_i$ are nonnegative integers.

Here $a_i$ stands for the number of dice that show face $i$.

Applying stars and bars we find $$\binom{n+s-1}{s-1}$$possibilities.

Answer 2

If order doesn't count, then consider the following setup (I will do it for three 6-sided dice, but it can easily be generalized):

Put up five dividers ("bars"): $$ |\quad|\quad|\quad|\quad| $$ These five bars give us six regions (four gaps between the bars, as well as to the left and to the right). Each region corresponds to a number from 1 to 6 (say from left to right). Now distribute 3 $*$'s ("stars") in these six regions in any way you want. Each star corresponds to the result of a die. For instance, a throw of $3, 3, 4$ looks like $$ |\quad|{}*{}*{}|{}*{}|\quad| $$ This way each die throw corresponds to one ordering of 3 stars and 5 bars, and each ordering of stars and bars corresponds to a die throw.

The ordering of $3$ stars and $5$ bars can be thought of this way: You have $8$ available spots for any of the two symbols, and you want to choose $3$ of those spost to contain a star (the remaining $5$ will get bars). That can be done in $\binom{8}{3} = 56$ ways.