The answer is : $6!/6^6$. but, i don't understand why the numerator is not one. because at the end of the day we just want the 6 dices to be different, so order doesn't matter.$1,2,3,4,5,6$ is the same thing as $6,5,2,3,1,4$ because they're all different and that's what we're looking for because we're throwing all at once.
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It's easier to think of this as constructively building the sequence of rolls, such that each roll must be different to each of the previous rolls.
1st Roll: There are no rolls done yet, so any number rolled must be new. Probability is 1.
2nd Roll: There is 1 number used, so the probability of a new number is 5/6. ...
so on, until the 6th role is 1/6.
If you multiply this out, you get $6!/6^6$.
Another explanation is possibly that since you are multiplying 1/6 for choosing an EXACT number on each dice (not just up to ordering), you need to compute the number of ways to pick a number on each of the 6 dice to get all different. This is $6!$.
Well, there’s only one combination of dice that has all the dice be different, but we’re not counting the probability that a random combination of dice is all different; we’re calculating the probability that a roll of dice will be this specific combination. For example, the probability of getting $5,5,5,5,5,5$ would be just $1/6^6$, because there’s only one roll of dice for that combination. But how many rolls of dice are there for $1,2,3,4,5,6$? The number of ways you can rearrange those, which is $6!$. Since the dice will roll all $6^6$ possible lists with equal likelihood, $6!/6^6$ percent of the dice rolls will be some permutation of $1,2,3,4,5,6$. If we actually just wanted the specific sequence $1,2,3,4,5,6$ it would be $1/6^6$, but since we don’t care a lot more rolls are valid, and there’s a lot more chances for it to be right.