One Dimensional Random Walk with Possibility to Stand Still I'm having some problems with the following question:

Consider a random walk in 1D where there's $\frac{p}{2}$ probability to walk to the right, $\frac{p}{2}$ probability to walk to the left and $q=1-p$ probability to doesn't move at all.
a) What's the probability to walk $n$ (of a total of $N$) steps to the right?

There are of course more items to answer, but all of them depend on this one. What I did so far was to arrive at the general formula
$\frac{N!}{N_1! N_2! N_3!} \left(\frac{p}{q}\right)^{N_1} \left(\frac{p}{q}\right)^{N_2} (1-p)^{N_3}$
but I've been trying for a while and I really can't find a way to write this formula as a function of the number os steps to the right $N_1$ and the total number of steps $N$.
Can someone give me a hand?
Thanks!
 A: I'm not sure if you are asking if the net position is exactly $n$ steps to the right of the starting position or if there are exactly $n$ right steps taken total.  I think it is the latter in which case you can reason as follows (assuming independent steps):
We have that for $l$ steps to the left, $r$ steps to the right and $s$ stationary "steps" ($l + r + s = N$, the probability for any given order of these steps is: 
$$ \left (\frac{p}{2} \right)^l \left (\frac{p}{2} \right)^r \left (1-p \right)^s$$
To account for all possible orderings we multiply by the multinomial coefficient $\binom{N}{l,r,s}$ and get:
$$Pr(L =l, R=r,S=s) = \binom{N}{l,r,s}\left (\frac{p}{2} \right)^l \left (\frac{p}{2} \right)^r \left (1-p \right)^s$$
This is exactly the Multinomial distribution.
Now, fixing $r = n$ we want the probability:
$$\sum_{l=0}^{N-n}\binom{N}{l,n,N-n-l}\left (\frac{p}{2}\right )^l\left (\frac{p}{2} \right)^n(1-p)^{N-n-l} $$ 
According to Wolfram Alpha this sum is:
$$\binom{N}{N-n}\left (\frac{(1-p)(p-2)}{2(p-1)} \right)^N \left (\frac{p(p-1)}{(1-p)(p-2)} \right)^n$$
I do not know a general method for solving such sums.
