If n balls are thrown into k bins, what is the probability that every bin gets at least one ball? If $n$ balls are thrown into $k$ bins (uniformly at random and independently), what is the probability that every bin gets at least one ball?  i.e. If we write $X$ for the number of empty bins, what is $P(X=0)$?

I was able to calculate the $E(X)$ and thus bound with Markov's inequality $P(X&gt;=1) \le E(X)$ but I don't how to work out an exact answer.

http://www.inference.phy.cam.ac.uk/mackay/itprnn/ps/588.596.pdf
What is the chance  that all $k$ bins are occupied?

For $1\leq i\leq k$, define $A_i$ to be the event that the $i$th bin stays empty.
These are exchangeable events with $P(A_1\cdots A_j)=(1-{j\over k})^n$ and so
by inclusion-exclusion, the probability that there are no empty bins is
$$P(X=0)=\sum_{j=0}^k (-1)^j {k\choose j}\left(1-{j\over k}\right)^n.$$

Stirling numbers of the second kind can be used to give 
an alternative solution to the occupancy
problem. We can fill all  $k$ bins as follows: partition the balls
$\{1,2,\dots, n\}$ into $k$ non-empty sets, then assign the bin values $1,2,\dots, k$ to
these sets. There are ${n\brace k}$ partitions, and for each partition $k!$ ways to 
assign the bin values. Thus, 
$$P(X=0)={{n\brace k}\,k!\over k^n}.$$
Set $X_i = 1$ if there is at least 1 ball in the $\textit{i}$th bin; and $0$ otherwise [$\textit{i}$ goes from 1 to k]. Then this question is asking what is $P[\sum\limits_{i=1}^k X_i = k]$. Given that each $X_i$s are independent of each other, $P[\sum\limits_{i=1}^k X_i=k]) = (P[X_i=1])^k =(1-P[X_i=0])^k =(1-(\frac{k-1}{k})^n)^k$