We can decompose the number 3 into the sum of positive integers in four different ways: $3 ; 2+1 ; 1+2$; and $1+1+1$. In the addition, the order of the terms matters, so $1+2$ and $2+1$ are considered different decompositions. How many different ways can we decompose the natural number $n$?
Consider the natural number $n$ as a chain consisting of $n$ links. To a decomposition of $n$ that meets the conditions of the problem, we assign that disconnection of the chain in which the pieces consist of the number of links corresponding to the terms of the sum.

|  |  | the second | between n |
| :---: | :---: | :---: | :---: |
|  |  | we do not cut | we cut |
| between the first two | we do not cut |  |  |
| between all others |  |  |  |

This is a one-to-one correspondence, which helps us easily count the possible decompositions of $n$. The chain can be disconnected at $n-1$ places, and at each place, we can choose between two options (i.e., whether to disconnect or not). Thus, the number of possible decompositions is $2^{n-1}$.