Counting sample points dice experiment 
Suppose that die have been altered so that the faces are $1,2,3,4,5,5$.
  If the die is tossed five times, what is the probability that the numbers recorded are $1,2,3,4,$ and $5$ in any order?

This exercise comes after a section titled Tools for Counting Sample Points.
In solving the examples they used this "method",

If a sample space contains $N$ equiprobable sample points and an event $A$ contains exactly $n_a$ sample points, it is easily seen that $$P(A)=\dfrac{n_a}{N}$$

My failed attempt:
$N=6^5$, since we are assigning $5$ rolls to 6 possible outcomes.
$n_a= P^6_5$, since assigning five rolls to the six outcomes will give us the event.
This answer is wrong.
Why is this wrong?
How should I approach this type of problem?
 A: There are six possible faces on which the die could land for each of the five rolls, so there are $6^5$ equally likely outcomes in the sample space.  
There are two ways to choose the face which shows a 5 in the sequence of rolls.  There are five ways to choose its position in the sequence of rolls.  The remaining four distinct outcomes can appear in the sequence in $4!$ orders.  Hence, the number of favorable cases is $2 \cdot 5 \cdot 4! = 2 \cdot 5!$.
Thus, the probability that the numbers 1, 2, 3, 4, 5 all appear in a sequence of five rolls of the die is 
$$\frac{2 \cdot 5!}{6^5}$$
In counting the favorable outcomes, you did not take into account the fact that two of the six outcomes are indistinguishable. 
A: Your problem is that the sample points are not equiprobable, rolling 5 is more probable
The probability that the numbers recorded are $1,2,3,4$ in any order is $\frac{4!}{6^4}$.
The probability to roll a $5$ the $n$-th time is $\frac{1}{3}$.
The probability to roll a $5$ the $n$-th time and rolling $1,2,3,4$ in any order otherwise is $\frac{1}{3}\cdot\frac{4!}{6^4}$.
There are five $n$-th times you could do that (first to fifth time), and because each of those $5$ events excludes the others, the probability for any of those events to happen is the sum of the probabilities of those events, which is $5\cdot\frac{1}{3}\cdot\frac{4!}{6^4}$
