$3$ dice are thrown simultaneously I have a doubt in this question:

Three dice are thrown simultaneously. Find the probability that:
  
  
*
  
*All show distinct faces
  
*Two of them show the same face

My approach is for 1):
$$
\frac{_6C_1}{_6C_1}\cdot\frac{_5C_1}{_6C_1}\cdot\frac{_4C_1}{_6C_1}\cdot 3!
$$
and for 2):
$$
\frac{_6C_1}{_6C_1}\cdot\frac{1}{_6C_1}\cdot\frac{_5C_1}{_6C_1}\cdot\frac{3!}{2!}
$$
but the correct solution is for 1): 
$$
\frac{_6C_1}{_6C_1}\cdot\frac{_5C_1}{_6C_1}\cdot\frac{_4C_1}{_6C_1}
$$
and for 2):
$$
\frac{_6C_1}{_6C_1}\cdot\frac{1}{_6C_1}\cdot\frac{_5C_1}{_6C_1}\cdot\frac{3!}{2!}
$$
 A: If you evaluate your answer to 1, you will find the probability exceeds 1, so it cannot be correct. Your factor of $3!$ is extraneous, as you didn't specify the faces, just that they be different from the ones that came before.
A: (a) Imagine that the dice are coloured blue, white, and red.
We look at the numbers on the dice, looking first at the blue, then at the white, then at the red.
It doesn't matter what is on the blue. The probability that the number on the white is different from the number on the blue is $\frac{5}{6}$. Given that the white was different from the blue, the probability that the red is different from both is $\frac{4}{6}$. 
It follows that the probability the numbers are all different is $\frac{5}{6}\frac{4}{6}$.
Equivalently, record the numbers on the blue, the white, and the red in that order. There are $6^3$ possible records. They are all equally likely.
There are $(6)(5)(4)$ records in which the numbers are all different. The required proability is therefore $\frac{(6)(5)(4)}{6^3}$. 
(b) A countng argument works nicely. As in the second solution to (a), record the result as an ordered triple $(a,b,c)$ where $a$ is the number on the blue, $b$ the number on the white, and $c$ the number on the red. Alternately, assume that the dice were tossed not simultaneously, but one after the other (it makes no difference to the probability). Record the result as $(a,b,c)$ where $a$ is the number on the first die tossed, $b$ the number on the second, and $c$ on the third. There are $6^3$ equally likely records.
Now many records have exactly $1$ of one kind and $2$ of another kind, like $(5,2,5)$?
The number we have $1$ of can be chosen in $6$ ways. For each of these ways, where the singleton occurs can be chosen in $3$ ways. And once we have done that, the kind we have $2$ of can be chosen in $5$ ways, for a total of $(6)(3)(5)$.
For the probability, divide by $6^3$. 
A: Color the dice red, green and blue.  
Part (a) All different  
The red die can show any value.  That leaves 5 out of 6 for the green die, and then 4 out of 6 for the blue die.  So $$P_{all\, different}=\frac{5}{6}\times \frac{4}{6}=\frac{5}{9}$$
Part (b)  
Consider the probability that all the dice show the same value.  The red die can land with any value, leaving only one value for the green die and the blue die.  So:$$P_{all\,same}=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$$
The only remaining possibility is a 2-match:$$P_{2-match}=1-\frac{5}{9}-\frac{1}{36}=\frac{15}{36}$$
