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?

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.