Go back to where we were before. We had five ways to pick the location of the 4. We then needed to pick the location of the two 2's from the four remaining rolls.
How many ways are there to pick two positions out of four?
It's not the probability. But it is the number of rolls of the form 422xy up to permutation, and that's enough.
now we use the fact that for each such roll, the probability is (1/6)^3*(4/6)^2
And since each such roll is exclusive---e.g. if you roll 422xy then you haven't rolled 242xy or anythign else---you just add up all of those probabilities
so that's a probability of (1/6)^3*(4/6)^2 for each of the $\binom{5}{1}\binom{4}{2}$ exclusive ways of get a useful roll
Can you put that together to get the final probability?
But, having specified the outcomes of the other two dice rolls, the probability is no longer (1/6)^3(4/6)^2. instead, it's just (1/6)^5 since each dice roll would have to be as you've specified
Because there's 5 choices for where the 4 is and 4-choose-2 choices for where the 2's are. (I'm not assuming anything about the x and y.)
However, my own way of working this out would be a bit different.
There are 6^5=7776 ways of rolling 5 dice in sequence.
If we want to create a dice roll with exactly two 2's and exactly one 4, then: there's 5 possible locations for the 5 and 4-choose-2 = 6 locations for the 2's. For the remaining 2 dice rolls, there's 4 choices each. So there are 5*6*4*4=480 ways to get exactly two 2's and exactly one 4.
So the probability is 480/7776, which is equivalent to what you wrote above.
