Find distribution of following r.v. There are $n$ different objects : $a_1 \dots a_n$. We may choose any objects and take it back. So everytime we have probability of choosing is $1/n$. 
Let's suppose we $n$ times choose some element. Now let's $\mathbb{P}(\xi = k)$ is probability that we choose exactly $k$ different objects. We want to describe this probability and find $\mathbb{E} \xi$.
My attempt : first of all we need choose $k$ elements among all. First element we choose with unit probability, next one with $(n-1)/n$ \dots the $k$-th one with $(n-k+1)/n$. And other our choices should be among this $k$ elements. Hence we have : $\mathbb{P}(\xi = k) = \binom{n}{k} \frac{n!}{(n-k)! n^k} \left(\frac{k}{n}\right)^{n - k + 1}$. 
Unfortunately I miss something because it even doesnt give me a distribution. Any hints?
 A: First let us find the probability that exactly the objects $a_{1},\dots,a_{k}$
are picked.
Define: $$\hat{S}:=\left\{ i\in\left\{ 1,\dots,n\right\} \mid\text{object }a_{i}\text{ is chosen}\right\} $$
Then $\hat{S}$ is a random set and if we state that $S:=\left\{ 1,\dots,k\right\} $
then by symmetry: $$P\left\{ \xi=k\right\} =\binom{n}{k}P\left(\hat{S}=S\right)$$
Further with inclusion/exclusion we find that: $$P\left(\hat{S}=S\right)=P\left(\hat{S}\subseteq S\right)-P\left(\bigcup_{s\in S}\left\{ \hat{S}\subseteq S-\left\{ s\right\} \right\} \right)=$$$$\sum_{T\subseteq S}P\left(\hat{S}\subseteq T\right)\left(-1\right)^{\left|S\right|-\left|T\right|}=\sum_{i=0}^{k}\binom{k}{i}\left(\frac{i}{n}\right)^{n}\left(-1\right)^{k-i}\tag1$$
So we end up with: $$P\left\{ \xi=k\right\} =\binom{n}{k}\sum_{i=0}^{k}\binom{k}{i}\left(\frac{i}{n}\right)^{n}\left(-1\right)^{k-i}$$

edit concerning the expectation of $\xi$.
In order to find $\mathbb E\xi$ you better not use the distribution of $\xi$.
It can be solved by applying linearity of expectation and symmetry.
For $i=1,\dots,n$ define random variable $A_i$ that takes value $1$ if object $a_i$ is chosen and takes value $0$ otherwise.
Then: $$\xi=\sum_{i=1}^nA_i$$ and consequently:$$\mathbb E\xi=\mathbb E\sum_{i=1}^nA_i=\sum_{i=1}^n\mathbb EA_i=nP(A_1=1)=n\left(1-\left(1-\frac1n\right)^n\right)$$

edit to enlighten the inclusion/exclusion part:
The inclusion/exclusion starts like this:
$$P\left(\bigcup_{s\in S}\left\{ \hat{S}\subseteq S-\left\{ s\right\} \right\} \right)=$$$$\sum_{s\in S}P\left(\left\{ \hat{S}\subseteq S-\left\{ s\right\} \right\} \right)-\sum_{s,t\in S,s\neq t}P\left(\left\{ \hat{S}\subseteq S-\left\{ s\right\} \right\} \cap\left\{ \hat{S}\subseteq S-\left\{ t\right\} \right\} \right)+\cdots$$
Note that we can write this also as:
$$P\left(\bigcup_{s\in S}\left\{ \hat{S}\subseteq S-\left\{ s\right\} \right\} \right)=$$$$\sum_{T\subseteq S,\left|T\right|=k-1}P\left( \hat{S}\subseteq T \right)-\sum_{T\subseteq S,\left|T\right|=k-2}P\left(\hat{S}\subseteq T\right)+\cdots$$
