First of all, I know there are a lot of these and similar questions already online but I am trying to solve this on my own and I need you to tell me what I am doing wrong. So, in class we solved problem like this: \begin{align*} \text{Probability} & = \frac{\text{positive events}}{\text{all possible events}}\\ & = \frac{\binom{4}{2} \cdot 5^2 + \binom{4}{3} \cdot 5 + 1}{6^4} \end{align*}
My question is why cannot we solve it like this?
$$\frac{\binom{4}{2} \cdot 6^2}{6^4}$$
where $\binom{4}{2}$ is the number of ways of obtaining six in two of the four trials and where $6^2$ are two remaining dice which can be any number but it does not matter which because we already have two six from choosing $2$ from $4$?
How many ways are there to roll 4 sixes?
Is it: choose 2 of the 4 to be sixes and it doesn't matter about the other 2 because they are also sixes? That is $\binom{4}{2}$, which is $6$ .
Or is it: $1$ ?
Therein lies the reason why. Your technique would over count such redundant results.