Show that the number of possible compositions for n ∈ $\mathbb{N}$ is $2^{n−1}$

Question

Let n ∈ $\mathbb{N}$ be a natural number, an ordered set of positive integers $(λ_{1},...,λ_{k})$ such that $λ_{1}\,+ \, ... \, + \, λ_{k} = n$ is called a composition for n ∈ $\mathbb{N}$. These integers are not necessarily distinct. Show that the number of possible compositions for n ∈ $\mathbb{N}$ is $2^{n−1}$.

Example: the number $n = 4$ has the following 8 compositions $(4),(3, 1),(2, 2),(2, 1, 1),(1, 3),(1, 2, 1),(1, 1, 2),(1, 1, 1, 1)$.

score 1 · Accepted Answer · 2019-01-18T00:49:00.167

1

By using stars and bars, the number of partitions into $k$ groups is $n-1\choose k-1$.

Adding up, and using the binomial theorem, we get $\sum_{k=1}^n{n-1\choose k-1}=\sum_{k=0}^{n-1}{n-1\choose k}=2^{n-1}$.

edited Jan 18 '19 at 00:49

answered Jan 18 '19 at 00:44

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Or more directly, in the unrestricted problem you have $n-1$ places to choose to place a bar or not. – Daniel Schepler Jan 18 '19 at 00:48
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@DanielSchepler very nice. – Jan 18 '19 at 00:52

Show that the number of possible compositions for n ∈ $\mathbb{N}$ is $2^{n−1}$

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