<s>[INST] 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.
 [/INST] Suppose we have a pair of indistinguishable white dice.  There are $\binom{6}{2}$ outcomes in which they show different numbers and $\binom{6}{1}$ outcomes in which they show the same number.  Hence, there are a total of
$$\binom{6}{2} + \binom{6}{1} = 21$$
outcomes when two indistinguishable dice are thrown.
Suppose we now paint one of the dice red.  If there are two different outcomes, there are two ways to do this since we can either paint the higher number or the lower number red.  If the outcomes are the same, it does not matter which die is painted red since both dice will still show the same number.  Thus, if we make the dice distinguishable, we obtain
$$2\binom{6}{2} + \binom{6}{1} = 36$$
possible outcomes as we should expect since there are six possible outcomes for each die.
Thus, the problem with your approach, as user525966 pointed out in the comments that it only makes sense to divide the number of outcomes on distinct dice by $2$ when the two outcomes are different.  If one die is red and the other is white, there are $6 \cdot 5 = 30$ such outcomes.  On the other hand, if the distinct dice show the same outcome, they still would show the same outcome if we painted both dice white.  Thus, if we start with the $36 = 6^2$ possible outcomes for two distinct dice, we can obtain the number of possible outcomes for indistinguishable dice by dividing the number of outcomes in which the two dice display different outcomes by $2$  and then add the number of outcomes in which both dice display the same number.
$$\frac{6 \cdot 5}{2} + 6 = \binom{6}{2} + \binom{6}{1} = 21$$
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