If we've got 10 coupons, what is expected number of different ones if there are 25 different types I can't figure out this problem :

There are 25 different types of coupon, all equally probable to get. If we have got 10 coupons, what is expected number of different coupons between them?

Solution is : E(x) = 8.38
 A: Using Stirling numbers we get
$$\frac{1}{25^{10}} \sum_{q=1}^{10} q\times
{25\choose q} \times {10\brace q} \times q!$$
which gives the expectation
$${\frac {31964050675249}{3814697265625}}
\approx 8.379184100.$$
What we  have done here  is choose the  $q$ different values  from the
$25$  different ones  where we  can  obviously represent  at most  ten
different values. We  then partition the $10$ slots  into $q$ sets and
choose a permutation of the $q$  that gives the value assigned to each
set. Finally multiply by $q$  because we are computing the expectation
of the number of different values which is precisely $q.$
Observe also that the probability of a coupon number $p$ not appearing
is $$\frac{24^{10}}{25^{10}}$$
which yields for the expectation
$$\sum_{p=1}^{25} \left(1-\frac{24^{10}}{25^{10}}\right)
= {\frac {31964050675249}{3814697265625}}$$
the same as before.
A: Let $X_i$ be an indicator random variable $=1$ if the $i^{th}$ coupon is present, and $=0$ otherwise.
Then $P[i^{th}$ coupon is present$] = [1 - (\frac{24}{25})^{10}]$
Now the expectation of an indicator r.v.  is just the probability of the event it indicates, so $E[X_i] = [1 - (\frac{24}{25})^{10}]$
By linearity of expectation we have expectation of sum = sum of expectations,
$E[\sum{(X_i)}] = \sum{E(X_i)} = 25[1 - (\frac{24}{25})^{10}] \approx 8.38$ 
