Your method is an arrangement. Using that formula you get the answer if the order mattered, but you clearly stated that it doesn't. Let me just try and give you an example:
1. Say you have the balls marked a, b, c, d and e. Then since the order doesn't matter, you get that case a b c = case a c b and therefore this case is not counted twice, but just once.
2. If you use your way of looking at things, the two cases outlined before are considered different, therefore each will be counted once, giving you twice as many options as compared to the first case.

Thank Dana,
however, I knwo that permutation is the case when to use it when the order does matter.
so in this case, why isn't arrangement possible?
I mean, can you give me the same example with slight changes that would illustrate the following?
1-a case when arrangement is used
2- a case when permutation is used
3- a case when combination is used.

Thank you so much for your help

Permutations are used when the number of balls doesn't matter... like for instance the number of cases for extracting whichever number of balls (1 to 5), but when the order does make a difference (meaning when a b c is not considered the same as b a c).
I'm sorry, I just cannot think of examples right now... I myself HATE permutations and probabilities... And I have not read an example of the sort in years. Maybe someone else can help you, or maybe the ball example is enough for you. I'm really sorry...