How many unit paths in $\mathbb R^3$ of length $9$ starting from the origin?

Question

I have a general formula for the number of unit paths to be $a+b+c \choose a$ (which I'm not even certain is correct). I know $a+b+c=9$ but how do you determine what the bottom should be? I believe the answer should be a summation but am not sure of what. Is it the summation of $9\choose k$ where $k=1...9$?

Answer 1

What are the points are reachable by unit paths of length $9$? They are precisely the points $(a, b, c)$ such that $a + b + c = 9$. Moreover, given a point $(a, b, c)$, we know how many unit paths there are from the origin to $(a, b, c)$. It is just an arrangement from the multiset containing $a$ "left arrows", $b$ "up arrows" and $c$ "diagonal arrows". This is just the multinomial coefficient $a + b + c \choose a \ b \ c$. So the answer is \begin{gather} \sum_{a + b + c = 9} {a + b + c \choose a \ b \ c}. \end{gather} Does this formula look familiar to you?