Giving it some further thought, I came up with this:
Since the constraint is that a red ball follows a green ball once or more, I will take one red ball and one green ball from the set and make a fixed group of them which will be treated as one item. n-2 balls that may be red, green or gray (depending on values of m, l, k) will remain and my set now has n-1 items. A permutation would then look something like x (g r) x x x x.
Looking at this, it appears that the item (g r) has n-1 possible positions and the remainder of balls may be arranged in (n-2)! ways. So it looks like the number of permutations that have a red ball following a green ball once or more is (n-1)(n-2)! = (n-1)!
So the probability I'm looking for can be calculated as (n-1)!/n! = 1/n (if m>0 and l>0; in cases where m=0 or l=0 the probability is 0). Could this be correct?