Given a $n$-sided fair die, show that the probability $P$ that all faces from $1$ to $11$ show when doing $3n$-rolls is bounded by
$1 - 11 \cdot \frac{(n-1)^{3n}}{n^{3n}} \leq P \leq 1 - 11 \cdot \frac{11(n-1)^{3n}}{n^{3n}} + 5 \cdot 11 \cdot \frac{(n-2)^{3n}}{n^{3n}}$
I am really stuck on this one. I figured that the exact probability is very hard to find and stirling numbers are involved. So the only chance I have is to figure out what events where used for that bounds. My guess is that the left one comes from the assumtion that (Edit: Warning this is not quite correct, check the comments)
$\mathbb{P}($Faces 1-11 don't occur at all$)=\frac{|E|}{|Q|}=\frac{11 \cdot (n-1)^{3n}}{n^{3n}}$
And if you look at the complement it is that all faces 1-11 at least occur once. Surely this is a lower bound but how could I get the upper bound, I hope that I did not overlook something very easy.