November 7th, 2006, 9:55 pm
QuoteOriginally posted by: sevvostThis simple question does not require recurrent equations - or any other calculations, for that matter. What is your point?You mean you have a probabilistic argument to solve it directly ? Well the point is that if you denote by f(n,k) such a number then f(n,k) can be decomposed in paths that go up at 1st step which is f(n-1,k) and paths that do right, f(n, k-1). There surely exists a way of counting directly the paths, I have to think about it, but if you already have one, I am interested in it. Edit:Ah, I may see.. the length of the path is n+k and you choose n point among them to turn right... the idea is to put this problem in bijection with a simpler problem. You're right. In very deed. Edit:Ok, clash , yes combinatorial or probabilistic means the same to me, my dictionnary doesn't translate the french 'dénombrement' correctly :/
Last edited by
iwanttobelieve on November 6th, 2006, 11:00 pm, edited 1 time in total.