Probability of an event with tossing a die 3 times One tosses a die three times. We denote the obtained result by
$(X, Y , Z)$.

$X \in \{ 1, 2, 3, 4, 5, 6\}.$
$Y \in\{ 1, 2, 3, 4, 5, 6\}.$
$Z \in\{ 1, 2, 3, 4, 5, 6\}.$

We define the event
$A = \text{“}X\leq 4$ and $Y\leq 4$ and $Z \leq 4$ and $X \neq Y \neq Z\text{”}$

Find $P(A)$.

I know the answer but I don't know how to get it.

$P(A)= 1/9$ with order
$P(A) = 4/56$ without order

It is about arrangement, permutation and combination.
I tried

$P(A)=3\times2\times1/6^3=1/36$

Can someone help me please?
Thank you.
 A: Like I mentioned the question is about arrangement and combination.
With order it is an arrangement;
Blockquote
$P(A)= \frac{Favorable Event}{All Possible Event}$
$P(A)= \frac{\frac{4!}{(4-3)!}}{6^3}= \frac{24}{216} =\frac{1}{9}$
Without order it is a combination;
$P(A)= \frac{\frac{4!}{(4-3)!*3!}}{6^3}= \frac{4}{216} =\frac{1}{54}$
A: With order there are $6^3$ total outcomes. 
Again in order, there are $4 \times 3 \times 2$ options, assuming you mean to say that $X,Y$ and $Z$ are all different from each other. Namely $4$ options for the first , $3$ for the second (as we must pick among $\{1,2,3,4\}$ minus the first value), and finally $2$ for $Z$.
In the ordered case the answer is $$\frac{4 \times 3 \times 2}{6 ^3} = \frac{4}{36} = \frac{1}{9}$$
as asked for.
I have no good idea about how you'd interpret "unordered" here ;the way it's formulated it's ordered by definition; I could add more if you would explain it better.
A: Let $E$ denote the event $X\neq Y\neq Z$ (this notation is unclear because $\neq$ is not a transitive relation).
Then $$P(A)=P(E\wedge \max(X,Y,Z)\leq4)=P(E\mid\max(X,Y,Z)\leq4)P(\max(X,Y,Z)\leq4)=$$$$P(E\mid\max(X,Y,Z)\leq4)\times\left(\frac46\right)^3$$
If $E$ is the event that $X,Y,Z$ take distinct values then $P(E\mid\max(X,Y,Z)\leq4)=\frac44\frac34\frac24$.
If $E$ is the event that $X\neq Y\wedge Y\neq Z$ then $P(E\mid\max(X,Y,Z)\leq4)=\frac44\frac34\frac34$.
A: For part 1, we can have the patterns
$$123,124,132,134,142,143$$
and repeat for X=2,3,4 for a total of $24$ positive outcomes.
There are $216$ outcomes in total, and $\frac{24}{216}=\frac19$.
