Independent Events where Order doesn't matter I want to know how to calculate the number of outcomes of an independent event that disregards outcome. This may sound very complicated or maybe I phrased it wrong, so I give the below example:
A die is rolled 3 times, so there are $6^3$ outcomes. However, they will be repeats. For example, {1,3,5}, {5,3,1} and {3,5,1} are all counted as different outcomes. I want the number of outcomes where the order doesn't count, so the 3 above outcomes are regarded as the same.
I tried $\frac{6^3}{3!}$, as $3!$ is the number of ways to rearrange 3 elements. However, since outcomes like {1,1,1} and {6,6,6} exist, the formula above doesn't count. So, is there are a formula for this?
 A: Yes, there is.  It is sometimes referred to as "multichoose" and gets the notation $\left(\!\binom{n}{r}\!\right)$ however this notation is not particularly consistent across multiple authors with some referring to the upper number as the number of balls and the bottom as the number of bins while other authors reverse that with the upper number as the number of bins etc...
Rephrasing your question in terms of balls and bins, if you have $a$ identical balls that you want to distribute among $b$ distinct bins, the number of arrangements will be:
$$\binom{a+b-1}{b-1}$$
This is equivalently written as $\binom{a+b-1}{a}$.  Be aware that some books and authors when presenting the formula will do so with $n$ playing the role of the number of balls while other authors might prefer $n$ to play the role of the number of bins instead, leading to some confusion as to which is the correct formula.
To remember the correct formula, it helps to understand where it comes from.  It can be proven using a technique referred to as stars-and-bars, arranging a sequence of stars and bars in a line with the number of stars in each gap between bars corresponding to the specific outcome.  You need one fewer bar than the number of bins to have the right number of gaps.  For your three six-sided dice example, the outcome of $\{1,3,5\}$ could be represented as $\star\mid\mid\star\mid\mid\star\mid$ for example while the outcome of $\{1,1,2\}$ is represented as $\star\star\mid\star\mid\mid\mid\mid$ etc...
It is worth reminding you that these outcomes that you are counting will not be equally likely to occur, the outcome of $\{1,2,3\}$ is six times as likely to occur as $\{1,1,1\}$ for instance if you were rolling dice (since it could have also occurred as $\{3,1,2\}, \{2,1,3\}$, etc...).
