List all ordered partitions of n

Haskell, 37 bytes

f 0=[[]]
f n=[a:x|a<-[1..n],x<-f$n-a]

xnor saved two bytes.

Python, 56 bytes

f=lambda n:[x+[n-i]for i in range(n)for x in f(i)]or[[]]

A recursive solution: The ordered partitions of n are a partition of some smaller i with 0<=i<n, followed by the remainder n-i as the last element. For a base case, n=0 only has the empty partition.

Python 2, 61 bytes

f=lambda n,s='1,':1/n*[eval(s)]or f(n-1,'1+'+s)+f(n-1,'1,'+s)

This isn't the shortest, but I really like the method because it's so weird.

Recursively generates and evaluates 2**(n-1) strings, like

1+1+1+1,
1,1+1+1,
1+1,1+1,
1,1,1+1,
1+1+1,1,
1,1+1,1,
1+1,1,1,
1,1,1,1,

for n=4. These strings evaluate to tuples representing all the partitions. Between any two 1's is either a +, joining them into a single number, or a ,, splitting adjacent sections.