Let us first look at a similar problem: how many paths are there from [tex](0,0)[/tex] to [tex](3,3)[/tex] if we can only move right and up?
Consider the image attached to the answer. The number in each square signifies the ways to reach that square.
At our starting square [tex](0,0)[/tex] (bottom left corner square), there is only one possibility of reaching there, by not doing anything.
Let us look at the square above [tex](0,0)[/tex], which is [tex](0,1)[/tex]. There is only one way to reach there. This is the same for all the squares at the bottom and the left side.
However, it gets more interesting for the remaining squares. At [tex](1,1)[/tex], there are two possibilities of reaching there: from [tex](0,0)[/tex] to [tex](0,1)[/tex] to [tex](1,1)[/tex] or from [tex](0,0)[/tex] to [tex](1,0)[/tex] to [tex](1,1)[/tex].
There is a quicker way to find the possibilities. Since the only way to reach a certain square is either from a square below or to the left, the possibilities of reaching a square are equal to the sum of the possibilities of reaching a square below it and the possibilities of reaching a square to the left of it.
For example, look at square [tex](2,3)[/tex], which has ten possibilities. This is because the square below it [tex](2,2)[/tex], has six possibilities while the square to the left, [tex](1,3)[/tex], has four.
Thus, for this problem, the answer is [tex]20[/tex].
Now, let us look at your problem, which is the 3d version of this problem. You can apply the same ideas, that the possibilities of reaching each square are equal to the sum of the possibilities of reaching the squares below, to the left, and backward.
It may take some time to get used to 3d (draw four 4-by-4 grids), but you should get the answer of [tex]1680[/tex].
Advanced
If you notice carefully, for the 2d problem, the grid forms Pascal's triangle. The [tex]r[/tex]th element of the [tex]n[/tex] row on a Pascal's triangle is given by [tex]\frac{n!}{r!(n-r!)}[/tex].
On our grid, the square with coordinates [tex](x,y)[/tex] will have a row number of [tex]x+y[/tex] and an element number of [tex]x[/tex].
Applying the formula for Pascal's triangle, we have [tex]\frac{(x+y)!}{x!y!} [/tex].
Substitute [tex]x=3,y=3[/tex] to get [tex]\frac{6!}{3!3!}=20[/tex].
You can look up Pascal's pyramid to solve the 3d version of the problem. The formula you should get is [tex]\frac{(x+y+z)!)}{x!y!z!}[/tex] for the point [tex](x,y,z)[/tex].
Substitute [tex]x=3,y=3,z=3[/tex] to get [tex]\frac{9!}{3!3!3!}=1680[/tex].
