I believe in the fact you are describing (it has been awhile since I took statistics), you would actually have to use a multinomial distribution as a binomial distribution assumes the "result" belongs to two outcomes (such as true/false on a quiz of 10 questions), whereas in the case of throwing a die, there are 6 possible outcomes for each of the ten throws. However, the multinomial distribution is what the call the "general" (or generic perhaps) form of the binomial, because you are plugging in "2" as the number of outcomes (and probably because it occurs so much in statistics work they wanted to give a two outcome multinomial distribution a name).
I don't know how to create equations but here is a link the formula and using it. Essentially you are taking the number of "trials" (dice throws in your case), divinding that 6!4! (which corresponds to you wanting six 3's and four 5's) and multiplying that by ((1/6)^6)((1/6)^4) (which can be of course simplified but might be easier as is to appreciate: 1/6 is the probability of getting a 3 or a 5, and you are raising it to the number of 3's and 5's that you want - 6 and 4). That number, is the probability of getting six 3's and four 5's.
Now, imagine a tree diagrim in your head of that problem. Not the entire tree but the general idea of it: you're going to have several arrangments of numbers right? All of the arrangments on their own have a probability of (1/6)^10 of being chosen: since 1/6 is the individual chance of getting a number on a dice, and (by the multiplication rule) you are doing that 10 times. So (1/6)^10 denotes the probability of getting any 10-dice combination. In the multinomial distribution equation, this probability of getting any 10-die combination of numbers, is the right side of the equation.
Now the left side, where we divide N! trials by 6!4! - that is the number of possible arrangements of 3's and 5's. Because in that tree diagram in your head - theres more than ONE way to get six 3's and four 5's right? If the goal is only to get 3's and 5's, regardless of when I get them in those 10 throws, I have many ways of going about doing that: thus, I need to know HOW MANY ways I can get six 3's and four 5's. Once I have that number, I simply multiply that by the probability of a specific 10-die combination of numbers - and viola that is the probability of getting six 3's and four 5's in 10 throws of a (fair) dice!