Q:

$n$ balls of $2^{n}-1$ colors, order not significant, how many combinations?

An example:
With $n = 3$,
We draw 3 balls. There are 7 different colors (or numbers). The order of balls does not matter, so [red, green, blue] is treated as being equal to [green, red, blue]. Colors could repeat, i.e. [gray, gray, gray] is a valid draw.
How do we calculate the total number of variations possible?

A:

If the number of balls of each colour is assumed to be greater than the number of balls that you draw out then this is a matter of combinations with unlimited repetition. We can consider the equivalent problem of deciding how many ways we can distribute $n$ balls into $2^{n} -1$ boxes as follows
$$\underline{\lVert\text{colour 1}\rVert\text{colour 2}\rVert\text{colour 3}\rVert \ldots \lVert\text{colour }2^{n}-1\rVert }$$
Where "colour 1" indicates that the balls in this box have colour 1, etc.
If we replace the representation of the separators by 1's and the representation of the number of balls in a box by a sequence of zeros (e.g. sequence starting 1001101... means that there are two balls of colour 1 chosen, no balls of colour two and one ball of colour 3 then we can consider the problem of counting the number of possible sequences of $n$ zeros and $2^{n} - 2$ ones. This is simply
$\displaystyle C(2^{n} + n - 2, n) =\frac{(2^{n} + n - 2)!}{(2^{n} -2)!n!}$
(this is read  "Choose $n$ items from a set of $2^{n}+n-2$ items")

