In this post, we will describe path counting using simple symmetric random walk in dimension 1. In this random walk there is equal probability to go forward or backward in one dimension.  When the walk starts from zero, we can describe it as sum of 1 to n i.i.d random variables each with a probability of 1/2.   This is a vector with values likely between -1,1 taken n times and each are equally likely. Since there are 2^n elements in {0,1} ^ n, we have the probability distribution as P(E1 =e1, E2=e2, ... En=en)  = 1/2 * 1/2 * 1/2 ... = as 1/(2^n)
We now say that the Events A that depend only on the first n i.i.ds are subsets of this space.
So if there are #A elements in this space, we have P(A) = #A/2^n.
Therefore, if we can count the number of elements of A, then we can compute its probability.
First, we assign to each initial position and to each sequence of n signs, a path of length n. We can construct a grid and trace this path as cartesian  coordinates.
Now let us take an example where we have an Event on this graph/grid  constituting of positions and we want to compute its probability.
Here is such an example : A = {S2 >= 0, S4 = -1, S6 < 0, S7 < 0}.
The paths comprising this event are shown below. We can count that there are six paths in A.
So P(A) = 6/2^6 = 6//128 = 3/64 which is approximately 0.047.
We have thus found probabilities by counting path.
Next we look at paths that start and end at specific points. The number of paths that start at one specific point (x,y) and end  at another specific point (m,n) is given by (m - x) Choose ((m -x + n - y)/2).
Additionally, If we take the start as (0,0) and finish at (n,i) then if (n+i) is not an even number, then there is no path that starts from (0,0) and ends at (n, i). Hence the totat number of paths = n Choose (n+i)/2 which is equal to zero.