If I flip 10 dice, what's the probability I get 1 1, 2 2s, 3 3s, and 4 4s?
Here is what I tried: I realized that we can model this question in terms of bit strings. A 0 can represent a single dice roll. Since we roll 10 dice, we have 10 0s. A 1 can represent a separator between different values. Since a dice has 6 possible outcomes, we need 5 1s to separate these values. For instance, the bit string
010010001000011
represents rolling 1 1, 2 2s, 3 3s, and 4 4s. There are ${10 + 6 - 1}\choose{10}$ ways of forming bit strings with 10 0s and 6 - 1 = 5 1s, and the above string is only 1 of these ways. Thus, the answer to the original question is $\frac{1}{{10 + 6 - 1}\choose{10}} = \frac{1}{3003}$.
However, my concern is that my example with the bit strings does not perfectly biject to the original question. Each bit string is equally likely to occur. However, are all combinations of rolls of the 30 dice equally likely to appear? For instance, would rolling 10 1s be less likely than rolling 5 1s and 5 2s since the latter has more permutations?