Question about Combinations and Flipping coins I'm having trouble getting to a formula on how to get the number of combinations from flipping a coin n times if the order doesn't matter...
I know that to get the total number of combinations is 2^n... So if a coin was flipped twice, combinations would be 2^2=4:


*

*Heads,Heads

*Heads,Tails

*Tails,Heads

*Tails,Tails


But what if I don't care about the order,so for example:


*

*Heads,Tails = Tails,Heads


Then I would have only 3 combinations.
I've come to the conclusion that the formula is n + 1 ... n being the number of times the coin is being flipped.
But what if it was a dice with 6 sides, or n sides... Would the formula remain the same or what would the formula be then?
 A: Suppose we have a multi set $\{\infty*x_1,\infty*x_2, \dots, \infty*x_k\}$.  That is, we have an unlimited supply of each of our $k$ elements. Then the number of $r-combinations$, which is the combinations (order doesn't matter) of length $r$, of this set is $r+k-1\choose r$.

To bring it back to your question, in the heads and tails example, $k=2$ because we have either heads or tails.  Additionally, we can have as many heads or tails as possible, that is they are not restricted by anything which is why there is an infinite supply of them.  Then, we just need to find the r-combinations of this where $r$ is the length of the sequences of values we want.  Plugging in the values, we get ${2+2-1\choose 2}=3$, so it holds in that case.
When we want to think of a dice, we just need to replace $r$ with the number of elements we want in our sequence. For a dice, $k=6$. 
If this interests you, look into multisets, combinations of multisets, and combinatorics in general.
A: For the coin scenario, if you flip $n$ coins then you can have any of $0, 1, 2, \ldots, n$ heads. It's pretty easy to see that there are $n+1$ different values there. (Although, as pointed out, they do not have equal probability of occurring.)
If you have $n$ $k$-sided dice, it's a little trickier to work out how many different combinations values you could get, but it's doable. In fact, it's an application of stars and bars, in particular the second version - how many ways can you divide $n$ rolls into $k$ different "bins" of possible values, given that some bins may contain no rolls? So based on that, there are ${n+k-1}\choose{n}$ ways to do it.
