Six dice, six different figures (model problem) I am thinking about the same problem as posted here: Probability of All Distinct Faces When Six Dice Are Rolled. 
If one rolls six dice, what is the probability of attaining six different figures. I do understand the solution but wondered why it is necessary to assume that the dice are distinguishable. If I assume they are not then I have exactly one desired result and 
$$
\binom{6 + 6 - 1}{6} = \binom{11}{6} = 462 
$$
possible results.
But I get a different probability in this case which leads me to the assumption that it is not working this way. But why exactly is that?
 A: You have used a correct Stars and Bars argument to show that there are $462$ ways to distribute $6$ objects in $6$ boxes. 
However, if we assume that the dice are fair, and do not influence each other, then these $462$ possibilities are not all equally likely.
Let us look at a much smaller example, two identical coins. There are $3$ ways to distribute these into $2$ boxes. However, in this case it is fairly clear that the probability of two heads is $\frac{1}{4}$ and not $\frac{1}{3}$. For whether the coins are distinguishable or not, it should make no difference to the probability if we toss the coins sequentially and not simultaneously. And sequential tossing tells us the probability is $\frac{1}{2}\cdot\frac{1}{2}$.
Roughly speaking, an extreme Stars and Bars case, such as all balls in the third box, has smaller probability than any specific more "even" distribution.
Remark: To see more informally that a probability model based on distinguishable dice is appropriate, assume that otherwise indistinguishable dice are made different by writing IDs on them with invisible ink. The writing should not affect the probability of events that do not involve the IDs, such as all six numbers being obtained. 
A: There are $6!$ combinations of six different numbers, and there are $6^6$ combinations of six rolls. The probability is then:
$$
p = \frac{6!}{6^6}
$$
