TITLE: Probability that in one rolling of 5 dice we obtain two '6' and one '5'?
QUESTION [5 upvotes]: First I made a mistake of not taking into account that this event is dependent. 
So to get two '$6$' the probability would be: 
$$ P_5(2) = \frac{5!}{2!(5-2)!} \cdot \left(\frac16\right)^2\cdot\left(\frac56\right)^3  $$
Where

$p = \dfrac16$
$q = \dfrac56$
$N= 5$ (the number of elements of the system)

Then there are $3$ dice left on the table and we want to know the probability that one of them is a '$5$'. We rule out the two dice with the '$6$' face on. So the number of element in the system is now $3$. So the probability would be:
$$ P_3(1) = \frac{3!}{1!(3-1)!} \cdot \left(\frac16\right)^1\cdot\left(\frac56\right)^2  $$
But then the professor told me that $p=\dfrac15$ and not $p=\dfrac16$ That's what I don't get.
Of course at the end we multiply those two probabilities to get the final probability that we want.

REPLY [5 votes]: But then the professor told me that $p=1/5$ and not $p=1/6$ That's what I don't get.

You have correctly evaluated the probability for the event that two from the five dice show 6. 
$$\mathsf P(N_6{=}2)=\binom 52 \dfrac{1^25^3}{6^5}$$
Now, of the three dice which don't show 6, each may show faces $\{1,2,3,4,5\}$ with equal probability.   Thus the conditional probability for a die showing 5, when given that it does not show 6, is $1/5$.   So the conditional probability for obtaining one 5 when given that two 6 have been obtained among the five dice is:
$$\mathsf P(N_5{=}1\mid N_6{=}2)=\binom 31\dfrac{1^14^2}{5^3}$$

Of course at the end we multiply those two probabilities to get the final probability that we want.

And the $5^3$ factors will cancel, so:
$$\mathsf P(N_6{=}2,N_5{=}1)=\binom{5}{2}\binom{3}{1}\dfrac{1^21^14^2}{6^5}$$
PS: This is called a multinomial distribution.