Three dice: conditional probability After throwing $3$ dice, we know that on every die there is a different number. What is the probability that there is a $6$ on exactly one dice?

I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then 
$$P(B)=\dfrac{6∗5∗4}{6^3}\text{ and }
P(A\cap B)=\dfrac{1*5*4}{6^3}$$
 A: The required probability is $$\frac {\binom 3 1 \times 5 \times 4} {6 \times 5 \times 4} = \frac 3 6 = \frac 1 2.$$
A: 
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then 
  $$P(B)=\dfrac{6∗5∗4}{6^3}\text{ and }
P(A\cap B)=\dfrac{1*5*4}{6^3}$$

Ah, no.  You are evaluating $A\cap B$ as the event of selecting a '6' on a particular die and two different faces for the other two dice. You want the event of selecting a '6' and two different faces for the three dice.
Let $A$ be the event that at least one of the die shows a '6'.
Let $B$ be the event that each die shows a distinct face.
So $A\cap B$ is the event of selecting a '6' and two values from the set of five remaining, while $B$ is the event of selecting three values from the set of six.
$$\begin{align}\mathsf P(A\mid B)&=\dfrac{\mathsf P(A\cap B)}{\mathsf P(B)}\\[1ex]&=\dfrac{\binom 52}{\binom 63}\\[1ex]&=\dfrac{(5\cdot 4)/2}{(6\cdot 5\cdot 4)/(3\cdot 2)}\\[1ex]&=\dfrac{1}{2}\end{align}$$
A: If I understood right from question and also from the discussion before. That every throw it is ensured that you get a different number (so all three different numbers). So you can get 6 only in one throw. 
Probability of getting 6 in first throw is $\dfrac{1}{6}$, so it is guaranteed that next two throws there will not be any 6. So the total probability in this case is $\dfrac{1}{6}.1.1 = \dfrac{1}{6}$.
Now for getting 6 in second throw; in the first throw the required probability is $\dfrac{5}{6}$, in the second throw the probability is $\dfrac{1}{5}$ and we do not bother about the third throw (as it is guaranteed that 6 will not be there). So the total probability in this case is $\dfrac{5}{6}.\dfrac{1}{5}.1 = \dfrac{1}{6}$.
For getting 6 in third throw we have, not getting a 6 in first throw is $\dfrac{5}{6}$, not getting a 6 in second throw is $\dfrac{4}{5}$ , getting a 6 in third throw is $\dfrac{1}{4}$, so total probability is $\dfrac{5}{6}.\dfrac{4}{5}.\dfrac{1}{4} = \dfrac{1}{6}$.
So finally total probability is $\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}= \dfrac{1}{2}$
