Final answer:
The total number of compositions of an integer n is 2(n-1) because there are n-1 places to decide whether to cut or not, each with two choices, resulting in 2(n-1) possible combinations.
Step-by-step explanation:
Proof of Total Number of Compositions of n Being 2(n-1):
A composition of an integer n is a way of writing n as the sum of a sequence of positive integers. The question is asking us to prove that the total number of compositions of an integer n is 2(n-1). The proof involves a bit of combinatorial reasoning.
Calculating Steps:
If we have an integer n, there are n-1 spaces between these units that we could potentially cut to create a composition partition. In each of these n-1 places, we have two choices: to make a cut or not to make a cut. This is binary, hence each place can be occupied by either '0' (no cut) or '1' (cut).
Therefore, for each of the n-1 places, having two choices results in 2(n-1) possible combinations of cuts. Each unique combination of cuts results in a unique composition of the number n, since the cuts divide the sequence of units into the summands that make up that composition.
This reasoning can be better understood by a simple example. For n=3, we have '1 1 1', and the two spaces can be either cut or not cut, creating the compositions '3', '2+1', '1+2', and '1+1+1'. Clearly, there are 2(3-1) = 22 = 4 compositions for n=3, which aligns with our rule.