Formula for how many possible end states any number of 6 sided dice throw can result in? I want to create a board game where you throw dice to determine where you land on a matrix of  tiles. The problem is I want to find out how many dice would be optimal for a fair chance of landing on any of the tiles.
So for one die, you have 6 possible end states: 1,2,3,4,5 and 6.
For two dice, I "calculated" 21 (I just listed them in Excel, because my math is not amazing, but I'm hoping my logic is solid enough): 1/1,1/2,1/3,1/4,1/5,1/6,2/2 etc.
I excluded 2/1, for example, because that would just be the second state reversed, which for our player would not make any difference.
For three dice, I got 52 end states, by ignoring the repeating states, like I mentioned for two dice.
So is there any formula I can use to determine if what I "calculated" above is correct and to help with higher number of dice (I'd hate to have to write down all the possible end states for 4 dice or above).
This is probably very easy to solve, but all the Internet returns when you search for "number of possibilities for three six sided dice" is 216 (6 cubed), which I'm guessing is all the possible states, taking the position of every die in consideration.
Thank you very much for your help :D
EDIT:
Thank you Arthur and David for your quick and especially comprehensive replies.
The bars and stars method seems really nice and the formula seems easy enough to use in Excel, if I need more dice.
Based on your replies, I even managed to find my 4 missing end states :)
The reason why I needed the end states only is because I wanted to see how quickly I can cover the matrix. It is indeed a chessboard-type matrix, with an area of 12x12 tiles (144 in total).
You can move starting in any of the corners by the value of any of the dice, then change direction once per each remaining die and move the value of them.
So let's say the player gets a 3,4,5 dice combination:
Starts from 1a and moves 5 tiles to the right, reaching 5a
from there, he goes 3 tiles down and reaches 5d
from there, he goes to the right 4 more tiles and reaches 9d, his final destination
What I've noticed is that I can fill up the entire board (meaning reach all the tiles at least once) with the first 18 end states (highest being 1,4,6).
Also, with only two random throws, I was able to reach a bit over 77% of the tiles, due to the fact that you can change direction for each die.
So even if the probabilities are significantly lower for hitting a 3,3,3, for example, you would still be able to reach a lot of tiles with a single dice combination.
Thank you so much for the quick and helpful replies. 
I hope I will be able to help somebody else as fast as you guys did :D
 A: Yes, there is a way. It's called "stars and bars" for reasons which will be obvious.
Imagine you have a long box with six chambers, one after the other, labelled 1 through 6, and you throw pebbles into it, noting which of the chambers it fell into. Assuming you're equally likely to hit any of the chambers, this is equivalent to throwing a die.
The point here is that we can now study the pebbles and the five dividers between the chambers. Let $*$ denote a pebble and $|$ a divider. Then a throw of 1/2/3 would be denoted as 
$$
*|*|*|\hphantom{{}*{}}|\hphantom{{}*{}}|
$$
while a throw of 3/3/3 would be denoted as
$$
|\hphantom{{}*{}}|*{}*{}*|\hphantom{{}*{}}|\hphantom{{}*{}}|
$$
What you can see here, is that we have translated "a result from throwing three dice" into "an ordering of three stars and five bars". Each possible ordering represents a single die result, and each die result is represented by some ordering. Thus what we want is the number of orderings.
There are 8 symbols in total. In other words, if we are to construct some arbitrary ordering, we have 8 spots, and we need to choose three of them to put down stars (and fill in the rest with bars). This can be done in $\binom83 = 56$ ways (seems like you missed a few, which isn't surprising). In general, the $8$ is the total number of symbols, which is 5 plus the number of dice (six chambers means five dividers, which means five bars), and the number of stars is the number of dice. So for four dice we get $\binom94 = 126$ possible cases.
A: I assume you mean that what numbers you get and how many times you get them are important, the order is not important.  For example, with $5$ dice,
$$2,2,2,6,6\quad\hbox{and}\quad 6,2,2,6,2$$
are the same, but
$$2,6,6,6,6$$
is different.  In that case the "stars and bars" technique shows that the number of possibilities from $n$ dice is
$$\binom{n+5}5\ .$$
If you are not familiar with the notation, this means
$$\frac{(n+5)(n+4)(n+3)(n+2)(n+1)}{5\times4\times3\times2\times1}\ .$$
This is $6$ if $n=1$, and $21$ if $n=2$, and $56$ if $n=3$.  I think you may have missed some possibilities for three dice.
But note as mentioned in comments that different outcomes will not have equal probabilities.  This may or may not be relevant to you.  If you want to give more details in your question you may get a more useful answer.
