I've been struggling with this for over an hour now and I still have no good results, the question is as follows:
What's the probability of getting all the numbers from $1$ to $6$ by rolling $10$ dice simultaneously?
Can you give any hints or solutions? This problem seems really simple but I feel like I'm blind to the solution.
The way I see this problem, I'd consider two finite sets, namely the set comprised by the $10$ dice (denoted by $\Theta$) and the set of all the possible outcomes (denoted by $\Omega$), in this particular situation, the six faces of the dice.
Therefore, the apparent ambiguity of the problem is significantly reduced by considering all possible mappings of the form $f:\Theta \mapsto \Omega$ that are surjective. Why exactly surjection ?
Recall that the definition of surjection implies that all values in the codomain must be hit at least once through the mapping of elements in the domain of the function. Under the circumstance, it's exactly what we're interested in, since we want to exclusively count all instances in which all the possible $6$ faces appear .
The number of surjective mappings can be found using the following identity :
$$m^n - \binom {m} {1}(m-1)^n +\binom {m} {2}(m-2)^n-\binom {m} {3}(m-3)^n + \cdots $$
(where $m$ denotes the number of elements in $\Omega$ and $n$ the number of elements in $\Theta$ ).
Let $A$ represent the happy event, given the facts I've presented you should be able to get: $$Pr(A)=0.27$$
Hint:
If $X_{i}$ denotes the number of dice that show face $i$ then it equals:
$1-P\left[X_{1}=0\vee X_{2}=0\vee X_{3}=0\vee X_{4}=0\vee X_{5}=0\vee X_{6}=0\right]$
Now apply inclusion/exclusion.
edit:
It leads to:$$1-\binom{6}{1}\left(\frac{5}{6}\right)^{10}+\binom{6}{2}\left(\frac{4}{6}\right)^{10}-\binom{6}{3}\left(\frac{3}{6}\right)^{10}+\binom{6}{4}\left(\frac{2}{6}\right)^{10}-\binom{6}{5}\left(\frac{1}{6}\right)^{10}=\frac{38045 }{139968}$$
