Different Face Dice - Permutations and Combinations If four fair dice are tossed, what is the probability that they will show four different faces?
Ok so I get that the answer is (1/6)^4 x 6P2, but am wondering why it is a permutation and not a combination? It seems that it shouldn't matter what the order is?
I guess a more general question would be that I am confused about the subtleties that lead to something being a perm or a comb and are there any tricks/general rules? Pretty much I'm just trying to do as many questions and hoping exposure will do the trick. Thanks for the help :) 
 A: Indeed in a sense order does not matter. Whether the dice are tossed one at a time or all at once does not affect the probability. Whether the dice are of different colours, or indistinguishable does not affect the probability. 
However, for the analysis it is very useful to imagine the dice are tossed one at a time, or that they are coloured $4$ different colours, and the outcomes are listed in (say) alphabetical order by colour. 
The issue with not considering order is that then the various possibilities for the dice, like all $6$'s, or two $6$'s, one $3$, and one $2$ are not all equally likely. So we cannot compute a probability by simply dividing the number of "favourables" by the total number of possible outcomes.
A simpler example may be useful. Imagine that we toss two dimes. We ask for the probability both land heads. We can perfectly well consider the possible outcomes to be zero heads, one head, two heads. That gives us a $3$-element sample space. 
However, for computing the probability of two heads, it is useful to imagine that the dimes are tossed one at a time. Then we get a sample space with $4$ equally likely outcomes, and we see at once that the probability is $\frac{1}{4}$.
