Let n > 0 be an integer. We are given a balance and n weights of weight 2^0, 2^1, . . . , 2^(n−1). We are to place each of the n weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed.
Determine the number of ways in which this can be done.
As expected from an IMO problem, very difficult! But interesting solutions at www.math.leidenuniv.nl/~desmit/pop/2011_imo_final6.pdf
Update (25 Aug 2011):
I did not write in clearly in the post. One of the solutions provided in the pdf is an oral 3 line solution to the problem. It cannot get smaller than this :P. Even if you have solved the problem, do have a look at the pdf