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. 


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 $z^{-2}$).
