On 28/05/2014 03:44, Fred Lunnon wrote:
<< Only if I were foolish enough to write them in terms of factorials. Why would I do that? >>
Because e(11, 3) = e(9, 7) = 66 , perhaps?
If so then my trinomial-coefficient formula is wrong, which would be no surprise at all, but I still don't see the motivation for writing either my (presumably wrong) answer or whatever the correct answer is in terms of factorials. (Binomial and multinomial coefficients are usually best left as they are rather than decomposed into products and quotients of factorials. So it seems to me, anyway. At least one way of writing the correct answer to a question of this sort will surely be as a multinomial coefficient, or a sum of a few of them, because each combination of so-many (2,1), so-many (1,2), etc., will give rise to a multinomial term. There will be only finitely many such terms because getting too far away from the "obvious" paths from (0,0) to the vicinity of (4N,0) or (3N,3N) will imply a path longer than the optimum. Handwave, handwave.) I should add that there might well be a proof that the shortest-path counts are equal that doesn't involve finding explicitly what they are. For instance, one might hope for a sort of eighth-turn operation that maps (2,1) to (1,2) to (-1,2) to (-2,1) etc. My brief attempts at a solution along these lines were not successful. -- g