Let's say you want to know the probability of getting at least 2/3 dice rolls to land on at least 16/20. How can this be represented using purely combinatorics?
Using somewhat fragmented knowledge of statistics, I am able to come up summing the two following probabilities:
Probability of getting 2 throws of 16+
$$\binom{3}{2}\times({\frac{1} {4}})^2\times(\frac34)^1=\frac{9}{64}$$
Probability of getting 3 throws of 16+
$$\binom{3}{3}\times({\frac{1}{4}})^3\times(\frac34)^0=\frac{1}{64}$$
Summed together
$$\frac{10}{64}=\frac{5}{32}$$
But it doesn't seem like a unified approach. Is there a way to solve the entire problem using combinatorics?
Your first approach leading to $\frac 5{32}$ is correct. The final formula yields the correct result, but I do not understand where it comes from or how to apply it to another problem. You are really considering that you are rolling a d4 and asking the chance you get at least two $4$s. The denominators evaluate to $4^3$, which is the correct number of possibilities for three rolls of a d4, but I don't see why you express the two denominators differently. The leading binomials and numerators make sense and are the same as in the previous calculation.
Sometimes it is easier to compute the probability of failing and subtract from $1$, but here there are two cases ($0$ and $1$ rolls of $4$) there, too, so it is no easier.