Probability in dice, Feller exercise I  am stuck with exercise 2 of Chapter 4 Feller vol 1 "an introduction to probability theory and its application".
Here I report the exercise text:
Five dice are thrown. Find the probability that at least three of them show
the same face.
The solution I tried is the following:
Considering the dice as indistinguishable we can map the problem as drawing 5 indistinguishable balls in 6 boxes.
In this way have that the size of the sample space (ball and sticks model) is 
$$
|\Omega|=\binom{10}{5}=252
$$
Then calling
$$A=\lbrace \text{at least 3 dice show the same face}\rbrace$$ 
I tried to comupute the size of $A$ in different ways:
1) If I set 3 dice as equal I have that the number of outcomes with 3 or more equal dice are
$$
|A|=\binom{8}{5}=56
$$
in this way the Probability of event $A$ results:
$$P(A)=\frac{|A|}{|\Omega|}=0.2\bar{2}$$
This result is slightly differen from the reported solution $0.21...$
Therefore I tried another way:
calling
$$\bar{A}=\lbrace \text{at max 2 dice show the same face}\rbrace$$
I computed
$$|\bar{A}|=\binom{6}{5}+\binom{6}{4}\cdot\binom{4}{1}+\binom{6}{3}\cdot\binom{3}{1}=126$$
where
$\binom{6}{5}$ corresponds to the number of configurations in which I have all the faces of the 5 dices different;
$\binom{6}{4}$ corresponds to the number of configurations in which I 2 dices equal and all the other faces different, the term $\binom{4}{1}$ is due to the fact that I can choose in 4 ways the box containing 2 balls;
$\binom{6}{3}$ corresponds to the number of configurations in which I have 2 dice equal, other two dice equal and one dice different from the other (3 occupied boxes) then the term $\binom{3}{1}$ is the number of ways in which I can choose the box containing 1 ball.
Whith this method I get 
$$P(A)=1-P(\bar{A})=0.5$$
I cannot understand where I make mistakes..
 A: The appropriate model has the five dice distinguishable. A more familiar example is tossing two coins. A model that treats two heads, two tails, and one of each as equally likely gives answers that do not match reality.
There are thus $6^5$ equally likely outcomes. We now need to count the favourables. We divide into cases.
(i)  All dice show the same face. That face can be chosen in $6$ ways.
(ii) Four dice show the same face, and one is different. The common face can be chosen in $6$ ways. For each such choice the location of the different face can be chosen in $\binom{5}{1}$ ways, and then its value in $5$ ways, for a total of $(6)\binom{5}{1}(5)$.
(iii) Three dice show the same face, and two are different from the majority face. The majority face can be chosen in $6$ ways, and the locations of the other two in $\binom{5}{2}$ ways. These locations can be filled with non-majority numbers in $5^2$ ways, for a total of $(6)\binom{5}{2}5^2$.
A: The 252 options are not equally likely.
One option has all dice rolling 1 - the probability is $(1/6)^5$.
Another option has one 1, one 2, one 3, one 4 and one 5.  But there are 120 ways this can happen, from 12345, 12354, 12435 and so on, depending on the value of die 1, which has 5 choices, and so on.  So the probability of getting all numbers 1 to 5 is $120/6^5$.
You don't get the 120 in the first case because die 1 has to give 1.
