Pascal Triangle - Ordered k-partitions

The number of integer solutions to
x1 + x2 + x3 = 6
with x1,x2,x3 >= 1 is equal to the number of integer solutions to
x1 + x2 + x3 = 3
with x1,x2,x3 >=0.
It is like distributing three similar balls into three different buckets, solutions are:
b b b | | ( x1=3, x2=0, x3=0 )
b b | b | ( x1=2, x2=1, x3=0 ).
A pattern appears. We are in fact looking at the number of different arrangements of three b's and 2 |'s, which is ( 3 + 2 )! / 3! 2! = C(5,2) = 10.
Let's list the solutions anyway.
3 - 0 - 0
2 - 1 - 0
2 - 0 - 1
1 - 2 - 0
1 - 1 - 1
1 - 0 - 2
0 - 3 - 0
0 - 2 - 1
0 - 1 - 2
0 - 0 - 3
or translated to the original question:

4 - 1 - 1
3 - 2 - 1
3 - 1 - 2
2 - 3 - 1
2 - 2 - 2
2 - 1 - 3
1 - 4 - 1
1 - 3 - 2
1 - 2 - 3
1 - 1 - 4.
They are in fact all the ordered 3-partitions of 6.
So we can interpret C(n,k) as the ordered (k+1)-partitions of (n+1).