Let's look at one possible way:
$$x_1, 5, x_2, 5, x_3$$
The $x_i$ stand for "anything else but 5." The probability of this event is $(\frac{1}{6})^2(\frac{5}{6})^3$. So, it can happen in this particular way OR (this is the key word) another way corresponding to _all the other ways to arrange the fives_. How many ways are there to arrange two objects in a set of five?
The answer is $\binom {5}{3}$. Hence, the probability of your event is $$\binom {5}{3}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3.$$