Conditional Probability with 6 dice

Question

Consider that a fair dice is thrown 6 times. I want to compute the probability that to have exactly two 6s "given" that we have exactly 2 5s. So I use the conditional probability equation: \begin{equation}\mathbb{P}(\text{two } 6s|\text{two }5s)=\frac{\mathbb{P}(\text{two } 6s\cap\text{two }5s)}{\mathbb{P}(\text{two }5s)}\end{equation} So the probability of throwing exactly two 5s from 6 throws is given by $15\times(\frac{1}{6})^{2}(\frac{5}{6})^{4}$, where the 15 comes from the 15 possible combinations to get exactly 2 5s. Now comes the part where I am stuck. Looking at the numerator, more precisly the intersection, I don't see what the intersection of the two events would be. Throwing 2 6s and 2 5s would be the union I believe. Is the intersection to throw 4 of 1,2,3 and 4? From that I believe I would also need to determine the number of combinations that we have this. Someone who knows how to compute my numerator?