I have a tree where, at each node, it splits into $b$ branches for a total number of $n$ levels.
I enumerate the paths from the root to the leaf nodes. For example, if $n = b = 2$ then I have the following paths from the root node to the leaf nodes:
{1,1}, {1,2}, {2,1}, {2,2}. (there are $n^b$ leaf nodes)
However, I'm only interested in paths that have unique sets, regardless of order. In the example I gave above, {1,2} and {2,1} are equivalent. I'm only interested in the elements.
For $n = 2, b = 3$, I have:
{1,1,1}, {1,1,2}, {1,2,1}, {1,2,2}, {2,1,1}, {2,1,2}, {2,2,1}, {2,2,2}
Where the bold sets represent the unique results.
Unless I've made a mistake in my enumeration, I've worked out the the first few examples by hand to try and get some insight into the problem:
$n$, $b$, unique results:
1, $k$, 1 (for arbitrary $k$)
2, 1, 2
2, 2, 3
2, 3, 4
3, 1, 3
3, 2, 6
3, 3, 10
How do I work out what the closed form is for the result I'm looking for given arbitrary $n$ and $b$?
You’re looking for the number of distinct multisets of cardinality $n$ whose elements all come from a set of cardinality $k$. This is sometimes written $\left(\!\!\left(k\atop n\right)\!\!\right)$ and is given by the formula
$$\left(\!\!\left(k\atop n\right)\!\!\right)=\binom{k+n-1}n=\binom{k+n-1}{k-1}\;.$$
The linked article has a reasonably clear derivation of the formula.