permutation and balls . we have 5 green balls 2 red balls and 3 blue balls we put them in a line next to each other what is the probability of having all the balls with the same color next to each other ?
does it matter if the balls are distinguishable or not ? i think the answer is no but why?
if we assume the balls are indistinguishable then 
the whole permutation of the balls is 
$ 10!/ (3! * 2! * 5! ) $  since the balls are the same.
and the desired permutation is $3!$
so the probability is $(3! * 5! * 2! * 3!)/10!$
if you calculate with distinguishable balls you get the same answer!
actually the real cause of my confusion is that sometimes in probability when we take objects as "indistinguishable" and solve the problem we get wrong results .
like when we want to calculate the probability of getting exactly one tail and one head when tossing two coins if we assume that the coins are "indistinguishable" then we miscalculate the probability as 1/3 instead of 1/2 ! 
how come sometimes the property of indistinguishability causes a problem and sometimes it don't ?  
in general when does distinguishability matter .
 A: The difference is that in the balls problem, each outcome in your sample space is equally likely in both the distinguishable and non-distinguishable cases. So, you can count both ways and get the same answer. However, in the coin toss case, the sample space for distinguishable coins is {HH, HT, TH, TT} all equally likely. So you can do $P\{1Head 1Tail\} = P(HT) + P(TH) = 1/4+1/4 = 1/2$ and get the correct answer. If you consider the coins indistinguishable, however, you would have sample space {2Heads, 2Tails, 1Head1Tail}, but the outcomes in the set are not equally likely. That's why it's wrong to use $1/3$. In general, if you want to use counting and division to calculate probability, make sure that each thing you count is equally likely to occur.
A: One thing is to count the number of different arrangements when  balls of the same color are indistinguishable, and another thing is to count arrangements with a particular property when $10$ colored balls are arranged at random.
(a) Assume that the $10$ balls are secretly marked and are arranged at random by a blind person. There are $10!$ arrangements.
(b) An arrangement is good if all balls of the same color are next to each other. There are six ways for the order of the three colors in a good arrangement. When such an order has been fixed, each good arrangement obeying this order corresponds to $5!\cdot 2!\cdot 3!$ arrangements when the $10$ balls are secretly marked as in (a). It follows that out of the $10!$ equally likely (a)-arrangements of the $10$ balls exactly  $6\cdot 5!\cdot 2!\cdot 3!$ are good. Therefore the probability $p$ that a random arrangement is good comes to
$$p={6\cdot 5!\cdot 2!\cdot 3!\over 10!}={1\over420}\ .$$
