Let $n>0$ be an integer. We are given a balance and $n$ weights of weight $2^{0}, 2^{1}, \ldots, 2^{n-1}$. In a sequence of $n$ moves we place all weights on the balance. In the first move we choose a weight and put it on the left pan. In each of the following moves we choose one of the remaining weights and we add it either to the left or to the right pan. Compute the number of ways in which we can perform these $n$ moves in such a way that the right pan is never heavier than the left pan.  Answer. The number $f(n)$ of ways of placing the $n$ weights is equal to the product of all odd positive integers less than or equal to $2 n-1$, i.e. $f(n)=(2 n-1) ! !=1 \cdot 3 \cdot 5 \cdot \ldots \cdot(2 n-1)$.
Assume $n \geq 2$. We claim  $$ f(n)=(2 n-1) f(n-1) . $$  Firstly, note that after the first move the left pan is always at least 1 heavier than the right one. Hence, any valid way of placing the $n$ weights on the scale gives rise, by not considering weight 1, to a valid way of placing the weights $2, 2^{2}, \ldots, 2^{n-1}$.  If we divide the weight of each weight by 2, the answer does not change. So these $n-1$ weights can be placed on the scale in $f(n-1)$ valid ways. Now we look at weight 1. If it is put on the scale in the first move, then it has to be placed on the left side, otherwise it can be placed either on the left or on the right side, because after the first move the difference between the weights on the left pan and the weights on the right pan is at least 2. Hence, there are exactly $2 n-1$ different ways of inserting weight 1 in each of the $f(n-1)$ valid sequences for the $n-1$ weights in order to get a valid sequence for the $n$ weights. This proves the claim.  Since $f(1)=1$, by induction we obtain for all positive integers $n$  $$ f(n)=(2 n-1) ! !=1 \cdot 3 \cdot 5 \cdot \ldots \cdot(2 n-1) $$  Comment 1. The word "compute" in the statement of the problem is probably too vague. An alternative but more artificial question might ask for the smallest $n$ for which the number of valid ways is divisible by 2011. In this case the answer would be 1006.  Comment 2. It is useful to remark that the answer is the same for any set of weights where each weight is heavier than the sum of the lighter ones. Indeed, in such cases the given condition is equivalent to asking that during the process the heaviest weight on the balance is always on the left pan.  Comment 3. Instead of considering the lightest weight, one may also consider the last weight put on the balance. If this weight is $2^{n-1}$ then it should be put on the left pan. Otherwise it may be put on any pan; the inequality would not be violated since at this moment the heaviest weight is already put onto the left pan. In view of the previous comment, in each of these $2 n-1$ cases the number of ways to place the previous weights is exactly $f(n-1)$, which yields (1).