How many ways from point A to point B? I am trying to solve this problem 
In how many ways can we go from point $A$ to point $B$ if we are only allowed to move along the arrows?
I use combinations and got an answer of $20$.
Since this is 3D grid
The height is $3$ units, length is $1$ unit and width is $1$ unit. Thus we can model the path to letters
$D$- downs,$L$-Left and $R$-Right
$$DDDLR$$
That is $5!$/$(3!*1!*1!)$
I would like to verify my answer. 

 A: Graphical method:
We can easily  calculate the  number of  different paths from $A$ to $B$ by adding the numbers of related sub-paths in the graphic.
                                                       

Algebraic method:
We introduce a coordinate  system and look for the number of different paths to go from $A=(0,0,0)$ to $B=(1,-1,-2)$ using steps $(1,0,0), (0,-1,0)$ and $(0,0,-1)$.
  
  
*
  
*We encode the step $(1,0,0)$ with $x$, $(0,-1,0)$ with $\frac{1}{y}$ and $(0,0,-1)$ with $\frac{1}{z}$.
  
*Each step has the form $x+\frac{1}{y}+\frac{1}{z}$.
  
*Since paths from $A$ to $B$ have length $4$ we are looking for the coefficient of $x\cdot\frac{1}{y}\cdot\frac{1}{z^2}$ in  $\left(x+\frac{1}{y}+\frac{1}{z}\right)^4$.
We use  the coefficient  of operator $[x^n]$ to denote the  coefficient of $x^n$ and obtain
  \begin{align*}
\color{blue}{[xy^{-1}z^{-2}]}&\color{blue}{\left(x+\frac{1}{y}+\frac{1}{z}\right)^4}\\
&=[xy^{-1}z^{-2}]\sum_{j=0}^4\binom{4}{j}x^j\left(\frac{1}{y}+\frac{1}{z}\right)^{4-j}\tag{1}\\
&=\binom{4}{1}[y^{-1}z^{-2}]\left(\frac{1}{y}+\frac{1}{z}\right)^{3}\tag{2}\\
&=\binom{4}{1}[y^{-1}z^{-2}]\sum_{j=0}^3\binom{3}{j}\left(\frac{1}{y}\right)^j\left(\frac{1}{z}\right)^{3-j}\tag{3}\\
&=\binom{4}{1}\binom{3}{1}\tag{4}\\
&\,\,\color{blue}{=12}
\end{align*}
in accordance with the graphical solution above.

Comment:


*

*In (1) we expand the binomial.

*In (2) we select the coefficient of $x^1$.

*In (3) we expand the binomial.

*In (4) we select the coefficient of $y^{-1}$ (which is also the coefficient of $z^{-2}$).
A: Assuming that the height is $3$ units as in the text and not $2$ as in the picture, then your result is correct.
Another way of seeing it but equivalent to your way is as follows:
Just put this in a 3D-coordinate system with 


*

*$B(0,0,0)$, $A(3,1,1)$
Each path from $B$ to $A$ corresponds to any arrangement of $\boxed{e_1 + e_2 + e_3 + e_3 + e_3}$ with $e_1 = (1,0,0), e_2 = (0,1,0), e_3 = (0,0,1)$.
There are $\frac{5!}{1!\cdot 1! \cdot 3!}=20$ different arrangements, hence, paths.
A: I get $12$ like others in the comments. One way to do this is to work backwards from $B$. All those points pointing into $B$ have no other outgoing arrows, meaning from each of those points there is only $1$ path. Then for all other points you add up the paths of all the points you could jump to next. So the points at middle height, left and right, both point into $2$ points with $1$ path, so there are $2$ paths from there to $B$. The same goes for the point in the back at the lowest level. So the point in the back at the middle level has $3$ arrows pointing out of it, each into a point with $2$ paths to $B$, meaning we get $2+2+2=6$ paths to $B$ for this point. At the upper level we get $1$ in the front, $3$ left and right and finally $3+3+6 =12$ at $A$.
A: If you want to avoid combinatorial argument (but I recommend Matti P. argument in the comments), just to do it step by step : 
First from $A$ you have two possibilities


*

*First going left (or right - there are the same by symmetry). From such a point it is immediate that there are only three possibilities : Down Down Left ; Down Left Down ; Left Down Dow. That's 6 possibilities

*If you decide to start by going down. Again do the same : Two choice Right/left or Down.


*

*If you go down, then you can only select left or right once. hence two paths.

*If you move left. Again you have only two paths : Down Left or Left Down. hence 4 paths on total (Two for left, two for right).



Hence a total of 12 paths. 
A: The discrepancy between 20 and 12 arises because there are only two down steps, not three.  The OP’s approach is correct for three down steps, but for two down steps the correct count is $$\frac{4!}{2!1!1!}=12.$$
