Mean value of taken unique cards' number The player takes cards from the deck with return. There are N cards and K attempts. K can be more or less than N. Probabilities to be taken for each card is equal. What is the mean amount of unique taken cards after K attempts of player?
I tried to solve this task by the next way:
The probability to take $K - i$ unique cards from the deck is 
$\frac{C_N^{K-i} (K - i)^i}{N^k}$ 
because there are $N^K$ ways to do $K$ choices and we can take $(K - i)$ different cards from deck by $C_N^{K - i}$ ways. Other $i$ choices can be only about cards that have been already chosen. It is $(K - i)^i$ ways.
In this way mean value is:
$M = \sum_{i=0}^{K - 1} \frac{C_N^{K-i} (K - i)^i}{N^k} (K - i)$
It is not right answer and it is not appropriate for the case when K > N. I have a suspicion that this task can be solved with multinomial distribution, but I can not understand how exactly I should do it.
 A: Classifying according to  the number $q$ of different  unique cards we
obtain  from  first  principles   for  the  number  of  configurations
(i.e.  number of  sequences of  draws  with replacement  from the  $N$
cards)
$${N\choose q} {K\brace q} q!$$
Let us verify that this sums to $N^K:$
$$\sum_{q=1}^N {N\choose q} {K\brace q} q!
= K! [z^K] \sum_{q=1}^N {N\choose q} (\exp(z)-1)^q.$$
We may add in $q=0$ as it does not contribute when $K\ge 1:$
$$K! [z^K] \sum_{q=0}^N {N\choose q} (\exp(z)-1)^q
= K! [z^K] \exp(Nz) = N^K.$$
The sanity check  goes through. Now for the expected  number of unique
cards we get
$$K! [z^K] \sum_{q=1}^N q {N\choose q} (\exp(z)-1)^q
= K! N [z^K] \sum_{q=1}^N  {N-1\choose q-1} (\exp(z)-1)^q
\\ = K! N [z^K] (\exp(z)-1)
\sum_{q=1}^N  {N-1\choose q-1} (\exp(z)-1)^{q-1}
\\ = K! N [z^K] (\exp(z)-1)
\sum_{q=0}^{N-1}  {N-1\choose q} (\exp(z)-1)^{q}
\\ = K! N [z^K] (\exp(z)-1) \exp((N-1)z)
= K! N [z^K] (\exp(Nz)-\exp((N-1)z))
\\ = N \times (N^K - (N-1)^K).$$
Divide by $N^K$ for the expectation
$$\bbox[5px,border:2px solid #00A000]{
N\times \left(1-\left(1-\frac{1}{N}\right)^K\right).}$$
Addendum. We  can also  answer the question  of what  the expected
number of singletons is. This is the species
$$\mathfrak{S}_{=N}
(\mathfrak{P}_{=0}(\mathcal{Z})
+\mathcal{U}\mathfrak{P}_{=1}(\mathcal{Z})
+\mathfrak{P}_{\ge 2}(\mathcal{Z})).$$
which yields the generating function
$$G(z, u) = (\exp(z)-z+uz)^N.$$
We get as a sanity check
$$K! [z^K] G(z, 1) = K! [z^K] \exp(Nz) =N^K$$
which is correct and the check goes through.
We obtain for the expectation
$$K! [z^K] \left.\frac{\partial}{\partial u} G(z, u)\right|_{u=1}
\\ = K! [z^K] \left. N (\exp(z)-z+uz)^{N-1} z \right|_{u=1}
\\ = K! N [z^{K-1}] \exp((N-1)z)
= K N (N-1)^{K-1}.$$
Divide by $N^K$ for the expectation
$$\bbox[5px,border:2px solid #00A000]{
K\times \left(1-\frac{1}{N}\right)^{K-1}.}$$
A: Assuming the question is "what is the expected number of values that are drawn at least once" we can answer via indicator variables, using linearity of expectation.
The probability that a given answer is observed is $$1-\left(\frac {N-1}N\right)^K$$  Summing over the $N$ possible values gives the expectation $$\boxed {N\times \left( 1-\left(\frac {N-1}N\right)^K\right)}$$
