In how many distinguishably different ways can a pair of indistinguishable dice come up? The answer is $21$, but why doesn't $(6\times 6)/2$ work? It accounts for the overlap where $2-4$ and $4-2$ are the same cases.

The critical point is they are $\color{blue}{\text{indistinguishable}}$ dice , so
Lets show the dice by stars and the values that dice can take is showed by gaps between bars. Then , we should have $5$ bars to satisfy $6$ gaps representing appearing values. For example , $||*||*|$ means that indistinguishable dice took the values of $(3,5)$. Another example is $|||||**$ means that indistinguishable dice took the values of $(6,6)$ . Another example is $*|||||*$ means that indistinguishable dice took the values of $(1,6)$ . Another example is $*|*||||$ means that indistinguishable dice took the values of $(1,2)$ . Do you get the idea ? We just permutate the stars and bars . The stars repesents the dice and the gaps between bars represent the appearing values on the dice
As you realize , we use stars and bars method for all possible permutation , so $$\binom{6+2-1}{2}=21$$ or $$\frac{7!}{5! \times 2!}=21$$
