Let n 0 be an integer. We are given a balance and n weights of weight 20,
21, . . . , 2n−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.

Question

Let n > 0 be an integer. We are given a balance and n weights of weight 20,
21, . . . , 2n−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.

anonymous · Answer

IMO Question Right?

anonymous · Answer

yeah

anonymous · Answer

I will try this after sometime...My head's not working now Lol ...

anonymous · Answer

kk

anonymous · Answer

This question doesn't make sense... what if n = 1?  It would go from 20, 21, ...?

dumbcow · Answer

Its an arithmetic series, 20,21,22,..20+(n-1)
The sum of 1st and last term will equal sum of 2nd and next to last term, and so on...
so if n is even, then the balance will be level after all the weights have been placed.