# Puzzles/Decision puzzles/Yet Another Weighing/Solution

< Puzzles | Decision puzzles | Yet Another WeighingPuzzles|Decision puzzles|Yet Another Weighing|Solution

One way to solve this would be to try to manually work out all the possible combinations and their sums. That would take a while.

Instead, notice that each item is half the weight of the previous item. Lets assume you have less items and work out the general rule for any number of items, where each is twice as much as the next item:

Items Weights Number of Combinations1 51: 5 2 5, 103: 5, 10, 15 3 5, 10, 207: 5, 10, 15, 20, 25, 30, 35 4 5, 10, 20, 4015: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75 5 5, 10, 20, 40, 80

Without even listing them, we can see the pattern. The numbers 1, 3, 7, 15 follow the formula of 2^{n} - 1, where n=1, 2, 3, 4 i.e. the number of items.

For 5 items, n=5, 2^{5} - 1 = **31**.

The pattern would continue for more items:

6 -637 -1278 -255...