Choosing a congress 
Five scientists each have to choose a venue to attend out of 8 possible venues. Their choice is independent and each venue has the same chance of being chosen.
  
  
*
  
*What is the probability that each of the 5 scientists choose a different venue?
  
*What is the probability that 4 out of 8 venues will not be chosen?

For the first one, it seems:
$$
\text{n} = \frac{8!}{3!}=8*7*6*5*4=6720\\
\text{N} = 8^5 =  32768\\
P(\text{choosing different venues}) = \frac{6720}{32768} =0.20507... = 20.5 \%
$$
The second one seems trickier. They are essentially saying: 4 venues are taken by 4 scientists. And the last guy should pick one of those. What is the probability of this?
A = 4 scientists pick 4 different venues
B = remaining scientist picks one of those 4
$P(A\cup B) = P(A) + P(B)$ because of the independence. So
$$
\frac{\dfrac{8!}{4!}+\dfrac{4!}{1!}}{32768}
$$
My gut feeling says that I’m off on the second question. Any hints or tips?
 A: Let us suppose we decide to use the sample space of $8^5$ that you used for the first part. Then for the numerator we want to find the number of choices in which exactly $4$ venues are chosen.
Which $4$? They can be chosen in $\binom{8}{4}$ ways. For every choice of $4$ venues, there are $\binom{4}{1}$ ways to choose the one that will be attended by $2$ scientists. Which $2$ scientists go to this popular venue? They can be chosen in $\binom{5}{2}$ ways. And once this is done, the remaining $3$ scientists can be distributed among the $3$ remaining chosen venues in $3!$ ways, for a total of $\binom{8}{4}\binom{4}{1}\binom{5}{2}3!$.
Or else we could choose the popular venue first, $\binom{8}{1}$ ways, and the scientists that go there, $\binom{5}{2}$ ways. That leaves $7$ venues, of which we choose $3$, and then multiply by $3!$. Same number.
A: Let us generalize and go out from the scenario that scientists show
up in some order and neatly on their turn choose a venue
out of $n$ venues numbered by $1,2,\ldots,n$. 
Denote the choice
of the $k$-th scientist by $C_{k}$. Then these are independent random
variables with $P\left\{ C_{k}=i\right\} =\frac{1}{n}$ for $i\in\left\{ 1,\ldots,n\right\} $.
For $k=0,1,\ldots$ denote $S_{k}=\left\{ C_{1},\ldots,C_{k}\right\}$
and $N_{k}=\left|S_{k}\right|$. Here $S_{0}=\emptyset$ and $N_{0}=0$.
For $k=1,\ldots$ and $r=1,\ldots,n$ we find:
$P\left\{ N_{k}=r\right\} =P\left\{ N_{k-1}=r\wedge C_{k}\in S_{r-1}\right\} +P\left\{ N_{k-1}=r-1\wedge C_{k}\notin S_{r-1}\right\} $ that is:
$P\left\{ N_{k}=r\right\} =P\left\{ N_{k-1}=r\right\} \frac{r}{n}+P\left\{ N_{k-1}=r-1\right\} \frac{n-r+1}{n}$
Denoting $p_{k,r}=P\left\{ N_{k}=r\right\} $ we have the trivials: $p_{0,0}=1$;
$p_{0,r}=0$ for $r=1,\ldots,n$ and $p_{k,0}=0$ for $k=1,\ldots$
Next to that there is the recursion relation just found:

$np_{k,r}=rp_{k-1,r}+\left(n-r+1\right)p_{k-1,r-1}$ for $k=1,2\ldots$
  and $r=1,\ldots,n$

You are asked to find $p_{5,5}$ and $p_{5,4}$ under the condition
that $n=8$. 
Applying the recursion results in:

$p_{5,1}:p_{5,2}:p_{5,3}:p_{5,4}:p_{5,5}=1:105:1050:2100:840$. 

Note here that $p_{5,5}=\frac{840}{4096}=\frac{6720}{32768}$ (agrees with
your answer on the first question) and $p_{5,4}=\frac{2100}{4096}=\frac{16800}{32768}$
(agrees with the answer of André on the second question).
Off course the answer of André is more direct, but it is not unthinkable that the recursion opens the way for a closed
formula for $p_{k,r}$ (or formally $p_{n,k,r}$ since it also is
depends on parameter $n$). However, I am not that far yet.
Edit
First fruit of recursion under parameter $n$: 

$E\left(N_{k}\right)=\frac{n-1}{n}E\left(N_{k-1}\right)+1$ so $E\left(N_{k}\right)=n\left(1-\left(\frac{n-1}{n}\right)^{k}\right)$

This can be achieved however on a simpler way. Let $Y_{i}=1$ if venue $i$ is chosen by some scientist and $Y_{i}=0$
otherwise. Then $E\left[Y_{i}\right]=1-\left(\frac{n-1}{n}\right)^{k}$
and $N_{k}=Y_{1}+\cdots+Y_{n}$ leading to $E\left[N_{k}\right]=E\left[Y_{1}\right]+\cdots E\left[Y_{n}\right]$
Defining $m_{n,k,r}=n^{k}p_{k,r}$ (corresponding with the number
of ways) we can translate the relation into:

$m_{n,k,r}=rm_{n,k-1,r}+\left(n-r+1\right)m_{n,k-1,r-1}$ for $n,k=1,2,\ldots$ and $r=1,\ldots,n$

Here $m_{n,0,0}=1$ and $m_{n,k,r}=0$ if $0\in\left\{ k,r\right\} \wedge k+r>0$. If I will not find a closed form for it then I will post a question about that myself.
