Probability of a die rolled three times yielding three even numbers A die is rolled three times. What is the probability of obtaining three
even numbers ?
I've solved this problem calculating the number of total results:
$$u=D'_{6,3}=6^3$$
and the number of favorable results:
$$f=D'_{3,3}=3^3$$
I've got:
$$p=\dfrac{f}{u}=\dfrac{3^3}{6^3}=\dfrac{1}{8}$$
This result is correct.
If i try to calculate $u$ and $f$ as combination with repetition, i get an error
Is it not possible to solve this problem with $u=C'_{6,3}$ and $f=C'_{3,3}$ ?
Why order of elements is so important ? The problem doesn't say anything about 
the order of the even numbers !
Thank you.
 A: Using combinations means you are choosing a certain number from a set, where order does not matter.
As for this problem, order does matter. $(2,4,6)$ and $(4,2,6)$ are different possibilities, therefore you should use permutations.
A: Toss 2 coins . What is the probability that it will result in $1$ head and $1$ tail? Note that the possible events can be $\{2H,1H1T,2T\}$
This shows why order is important in probability.
A: Since the question itself showed a correct way to find that the probability
of three even numbers is $\frac 18$, 
I'll look instead at this part of the question:

Why order of elements is so important? The problem doesn't say anything about the order of the even numbers!

The problem specifies that you have three numbers each resulting from a roll
of a fair die. Indeed it says nothing about the sequence of those numbers,
and if it is convenient for you to ignore the order of the numbers then
you may do so ... provided you also avoid the pitfalls described below.
The three dice will come up with certain combinations of numbers
with fixed frequencies regardless of who is asking what question about
those numbers. For example, 
there is a probability $\frac{1}{216}$
that all three dice will show the number $2$,
and the probability to get one of each of the
even numbers (one $2$, one $4$, and one $6$) in any order is $\frac{1}{36}$,
six times greater than the probability to get all $2$s.
But if you only count combinations of numbers selected
from $\{1,2,3,4,5,6\}$, you will effectively assign zero probability
to the even that all three dice show $2$.
Even if you decide to consider the order of the dice and count permutations,
you will not count all the events correctly.
And if you come up with some other counting scheme in which 
"all $2$s" is one event and "one each of $2$, $4$, and $6$" is another event,
you must multiply each event by its probability of occurring, 
not just count the events.
The formula "number of successes / total number of events" is valid only
when all events are equally likely.
In summary, it is not necessary to consider the order in which the numbers
are rolled when computing your probability, but it makes it much easier
to compute the correct answer.
A: Here's another way to think about it.
Your die throws are independent events (the value on the second or third throw doesn't depend on the first).
The probability of getting an even number is $1/2$ on each throw.  So getting an even number on all three throws is $1/2^3 = 1/8$.
What matters in your solution is that you counted the possibilities correctly.  There are indeed $3^3$ ways to get three even numbers, and $3^6$ total possibilities.
