Hi,
The number for 5 balls is 2^4 or 16, not 13. Below you omitted these 3 ways to
arrange the balls into groups of 3.  (1,2,2), (2,1,2), (2,2,1). You only gave
these three: (1,1,3), (3,1,1), (1,3,1). See below where I put in those missing
three:
 
Here is the reasoning.  First think of the 5 balls lined up in a row like this:
 
o o o o o 
 
Now notice that in between the 5 balls there are 4 places to put (or not put) a
partition "|", indicated by the 4 "?"'s below:
 
o?o?o?o?o
 
[I'll put your words between >> and <<] 
 
Here's the explanation. Follow me carefully:
 
>>I have 5 balls. I can arrange the balls in 1 group of 5 balls (5)<<

Yes, that's C(4,0) = 1 way because you are inserting no, that's 0, (zero),
partitions to separate the balls, so you have 1 set of 5 balls:
 
o   o   o   o   o
 
>>If I divide the 5 balls in two groups they can be arranged (1,4) (4,1) (2,3)
(3,2)<< 
 
Yes, that's C(4,1) = 4 groups of 2 because you are inserting 1 partition to
separate the balls
 
o|o o o o = (1,4)
o o|o o o = (2,3) 
o o o|o o = (3,2)
o o o o|o = (4,1)

If I divide the 5 balls in three groups they can be arranged (1,1,3) (3,1,1)
(1,3,1) (1,2,2) (2,1,2), (2,2,1) [putting in the 3 that you omitted] 

 
Yes, that's C(4,2) = 6 because you are inserting 2 partition to separate the
balls
o|o|o o o = (1,1,3)
o|o o|o o = (1,2,2)
o|o o o|o = (1,3,1)
o o|o|o o = (2,1,2)
o o|o o|o = (2,2,1)
o o o|o|o = (3,1,1)
 
>> If I divide the 5 balls in four groups they can be arranged (1,1,1,2)
(2,1,1,1) (1,2,1,1) (1,1,2,1)<< 
 
Yes, that's C(4,3) = 4 because you are inserting 3 partition to separate the
balls.

o|o|o|o o = (1,1,1,2)
o|o|o o|o = (1,1,2,1) 
o|o o|o|o = (1,2,1,1)
o o|o|o|o = (2,1,1,1)
 
>>If I divide the 5 balls in five groups they can be arranged (1,1,1,1,1)<<
 
Yes, that's C(4,4) = 1 because you are inserting 4 partition to separate the
balls
o|o|o|o|o = (1,1,1,1,1)
 
>>Always the number of balls used is five.
Total number of arrangements = 13<<  

No, it's 16

>>How can I calculate the number of arrangements for n number of balls ?<<

 
Thanks<<
 
The reason the total answer is 2^4 16 is because
 
o?o?o?o?o
 
Think of each question mark as asking the question: 'Does a partition "|" go
here?" There are 2 possible answers to that question "YES" and "NO". So the
number of ways to insert partitions "|" is obviously 2^4 or 16.
 
Notice also by the binomial theorem that
 
2^4 = (1+1)^4 = C(4,0)1^4 + C(4,1)1³1+C(4,2)1²1²+C(4,3)1*1³+C(4,4)1^4 =
 
C(4,0)+C(4,1)+C(4,2)+C(4,3)+C(4,4) = 1+4+6+4+1 = 16
 
 
If you have n balls there are n-1 places to put partitions "|", just like above
when you had 5 balls, there were 5-1 or 4 places to put partitions "|".
 
If you divide the n balls into m groups they can be arranged in C(n-1,m) where
m=0,1,...,n-1
 
But the total is simply 2^(n-1), why?  Same reasoning as with 5 balls. Suppose
you have n balls:
 
o?o?o?o?o...o?o
 
and n-1 question marks.  Think of each question mark as asking the question: 
'Does a partition "|" go here?"  There are 2 possible answers to that question,
"YES" and "NO".  So the number of ways to insert partitions "|" is obviously 
2^(n-1).
 
I think that answers your question.
 
Edwin