I am reading about a problem that states the following:
Suppose you have a balance and are allowed to choose the weights for its functionality. The objective is to pick the weights in a way that maximizes the number of items we can check by the scale. Given $n$ weights all of which are different how many different ways are there to arrange them on the scales of the balance? Assuming that the two sides of the balance are indistinguishable what conclusion can we make about the max range of items that we can use the scale to weigh?
The solution logic goes as follows:
Each weight can be placed on the left scale, the right scale or on the table. These are 3 options so for $n$ weights we have $3^n$ configurations of the balance. One of these is symmetric between left and right scale (when no weights are on the scales). The rest are asymmetric between the left and right scale. Which means the same configuration is counted twice.
So far it is clear. So we would have for e.g. ${1gram, 2gram}$:
Left, Right, Table
- - (1,2) (--)
1 - (2)
- 1 (2) (*)
2 - (1)
- 2 (1) (*)
1 2 (-)
2 1 (-) (*)
1,2 - (-)
- 1,2 (-) (*)
So we would have all the marked with $*$ as duplicates and the $--$ i.e. all on the table as the symmetric one we exclude since no weights are on the scales.
But then it states:
The total number of different configurations is $\frac{3^n - 1}2 + 1$
Where is the $+1$ coming from?
Going over an example for e.g. $n = 2$ for example $1gram, 3gram$
We can measure items of ${1gram, 2gram, 3gram, 4gram}$ which is a total of $\frac{3^2 - 1}2 =4$ and not $5$ as the solution claims.
Am I misunderstanding something here?
The +1 term comes from weighing zero grams, where there are no weights on either side.