Which is more likely with perfect six-side dice? Find the probability of the following 
$(i)$ to roll a $1$,then a $2$ and then a $3$ on consecutive roll of a single dice,
$(ii)$to roll three identical dice and have the outcome be a $1$,a $2$ and a $3$.
Which is more likely with perfect six-side dice?

My Attempt:
Probability to roll a $1$,then a $2$ and then a $3$ on consecutive roll of a single dice$=\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}+\frac{5}{6}\times(\frac{1}{6})^3+(\frac{5}{6})^2\times(\frac{1}{6})^3+(\frac{5}{6})^3\times(\frac{1}{6})^3+.....$
$=\frac{1}{36}$
Probability to roll three identical dice and have the outcome be a $1$,a $2$ and a $3=\frac{6}{216}=\frac{1}{36}$
There are 216 sample points in the sample space when we throw three dices and record the outcomes.Out of them six are favourable.$(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)$
So the two events are equally likely.
But the book answer says $(ii)$ event is more likely.I dont know how.Is some mistake i have made.Please help me.

Let's rephrase the question:
Which is more likely with perfect six-sided dice?


*

*Roll $1$ die $3$ times and get 1, then 2, then 3

*Roll $3$ dice $1$ time and get 1, 2, 3


Obviously, without even computing the probabilities, the second option is more likely.
This is because in the first option the order matters and in the second option it doesn't.

That said, let's compute the probabilities:

Roll $1$ die $3$ times and get 1, then 2, then 3:

$$\frac{1}{6^3}=\frac{1}{216}$$

Roll $3$ dice $1$ time and get 1, 2, 3:

$$\frac{3!}{6^3}=\frac{6}{216}$$
