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So I have to show that the number of partitions of $n \in \mathbb{N}$ is $2^{n-1} -1 $. ( the order is important ).

So here is my attempt. Please be strict. If you find any mistake or something that iritates you, let me know:

We rewrite $n \in \mathbb{N}$:

$n = 1 + \cdots + 1 $ .( we sum $1$ $n$-times). We want to find the number of compositions with at least two parts, so we consider "$+$". For each "$+$" between the summands we have to decide if the summands are a "new" summand or not. We get $2^{n-1}$ decisions. But we don't count the solution $n = n$ , so we have $2^{n-1} -1$.

What do you say? Remark: Please focus on my work. Thank you in advance.

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    $\begingroup$ That's not what I call partitions. $\endgroup$ – Lord Shark the Unknown Oct 22 '18 at 17:23
  • $\begingroup$ ok. any other suggestions? $\endgroup$ – Memories Oct 22 '18 at 17:25
  • $\begingroup$ Partitions where order matters are usually called "compositions". In addition $n$ would usually be considered a composition of itself, so you're looking to count compositions of $n$ with at least two parts. $\endgroup$ – Michael Lugo Oct 22 '18 at 17:26
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The wikipedia page https://en.wikipedia.org/wiki/Composition_(combinatorics) confirms both your answer and your argument.

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