In how many ways can i partition a stick I have a stick (or a ruler) of 10 cm length.
I want to cut the stick into pieces.
Each cm (1 cm, 2cm, 3cm, ... 0 cm) is a possible cutting point.
In how many different ways can i cut the stick?
One way would be to cut at every cm, so I get 10 parts.
Another would be to cut two 1cm pieces, and the rest 2cm pieces = 6 parts
Is there a formula for this? All i know is that it's not a normal permutation...
edit:
A 1cm piece cut at 1cm is different from a 1cm piece cut from 9cm, even when the rest is partitioned the same way.
 A: I do not think this is a partitions problems. In partitions the order of the summands do not matter. That is $1 + 8 + 1 = 10$ and $8 + 1 + 1 = 10$ are counted as the same partition, but they correspond to different ways of cutting the ruler. The first $1 + 8 + 1 = 10$ corresponds to cutting a 1cm and 9cm while the second $8 + 1 + 1 = 10$ corresponds to cutting a 8cm and 9cm. The questions as posed is how many ways to cut the ruler, this is actually a compositions problem. Like Mr.Wizard said the solution is $2^9 = 512$. This is because of each of 1cm, 2cm, ..., 9cm there can either be a cut or not. Thus we have two options (to either cut or not) and we do this 9 times. Therefore we get $2^9$.
See below for more on compositions. Compositions are a much easier problem than partitions.
https://en.wikipedia.org/wiki/Composition_(combinatorics)
A: There are two interpretations of your problem: if cutting into $[0,1]$ and $[1,10]$ is different than cutting into $[0,9]$ and $[9,10]$, then as was mentioned in the comments this is a counting principle problem: you decide independently which of nine points to cut or not cut at, in the same way as given five shirts, three pairs of shoes, and two pairs of trousers you have thirty choices of outfits.
But what seems to be a more natural interpretation is that the two cuttings I mentioned are the same, in which case this is a partition problem, the general type being to determine how many ways we can write an integer $n$ as a sum of smaller integers. Your problem is a partition problem because the pieces of ruler you come out with must have integer lengths summing to 10.
The easiest way to compute partition numbers by hand is to use Euler's pentagonal number theorem, which is described on the Wikipedia page. It's also possible, though it'll probably take a little while, to compute this partition number directly: you have partitions $10,9+1,8+2,8+1+1,7+3,7+2+1,7+1+1+1,6+4,...,5+4+1,...,5+1+1+1+1+1,4+4+2,...,1+1+1+1+1+1+1+1+1+1$.
The thing you must be careful about is not to list any partitions twice: this is why I wrote them down with pieces in decreasing order, which will guarantee you don't write, say, both $6+4$ and $4+6$. You should end up with the answer to life, the universe, and everything, so it's surely a worthwhile problem.
