How would I be able to use combinations or permutations to show double dice probability? I am doing an experiment where I am comparing the probability of each of the numbers from $2$ to $12$ when using two dice. Getting the percentage for each number is easy enough, but now I want to show how many possible combinations of dice rolls are possible for each number. I would preferably like to use combinations/permutations to show this, but got stuck in process of putting everything into the equations.
We have the formula $$C(n,r)=\frac{n!}{r!(n-r)!}$$ for combinations, but would $n=36$ since there are $36$ total combinations when using double dice, or would it be $12$ because there are $12$ possible numbers that can be rolled?
Thank you!
 A: This is not a problem which should have any mention of combinations or permutations.  Many problems in combinatorics have nothing to do with either and instead rely on more fundamental principles of counting.
As for how to actually approach, the specific example of two fair six-sided dice is a common every-day one which many people just learn out of practice in using them in common boardgames.  There are six different ways to get a sum of $7$ (1-6, 2-5,... 5-2, 6-1).  From there, moving to the left or to the right will decrease the amount by one for each step, so there are five ways to get a sum of $8$, five ways to get a sum of $6$, four ways to get a sum of $9$, etc... on to just one way to get a sum of $12$ and one way to get a sum of $2$.
If you insist on referring to a formula to count with... then I would recommend referring to the generating function for this specific scenario.
$$(x+x^2+x^3+x^4+x^5+x^6)^2$$
which expands as:
$$x^2+2x^3+3x^4+4x^5+5x^6+6x^7+5x^8+4x^9+3x^{10}+2x^{11}+x^{12}$$
The number of ways to get a sum of $k$ is the coefficient of $x^k$ in the expansion of the above.
The probability of getting one of these results is the corresponding coefficient divided by $36$ in this case, so getting a sum of $6$ for instance occurs with probability $\frac{5}{36}$.  When talking about getting one of the sums from a collection of possible sums... just simply add the totals (or probabilities) together.  For example, to get a sum of $6,10$ or $12$ occurs with probability $\frac{5}{36}+\frac{3}{36}+\frac{1}{36}$.
In some games, they even go so far as to have a handy reminder of these values, such as in the case of number chips for Settlers of Catan (pictured below) where the number of ways in which a sum can occur is represented by the number of dots on the chip.

This generating function easily generalizes.  Given a total of $d$ fair dice with $s$ sides, the sides being labeled $\{1,2,3,\dots,s\}$, the number of ways to get a total of $k$ will be the coefficient of $x^k$ in the expansion of:
$$(x+x^2+x^3+\dots+x^s)^d$$
and the probability will be that corresponding coefficient divided by $s^d$
