The names of 8 students are listed randomly. What is the probability that Al, Bob, and Charlie are all in the top 4? My attempt at a solution: There are $C_{4,3}$ ways to arrange Al, Bob, and Charlie in the top 4, and $C_{5,4}$ ways to arrange the other 5 people. All of this is over $C_{8,4}$, giving $$\frac{C_{4,3}C_{5,4}}{C_{8,4}}=\frac{2}{7}$$ Is this correct?
 A: No. 
There are $C_{8,3}$ arrangements possible for Al, Bob and Charlie, all having equal probability to occur.
In $C_{4,3}$ of these arrangements Al, Bob and Charlie end up in the top $4$.
So the correct answer is: $$\frac{C_{4,3}}{C_{8,3}}=\frac1{14}$$
A: Here is a slightly different approach:
There are $\dbinom{8}{4}$ ways to select the top 4, and
there are $\dbinom{5}{1}$ ways to choose the top 4 including Al, Bob, and Charlie 
$\;\;\;$(since 1 person out of the remaining 5 must be selected),
so the probability is $\displaystyle\frac{5}{\binom{8}{4}}=\frac{5}{70}=\frac{1}{14}$.

Alternate solution:
There are $8!$ ways to list the 8 people in order, and
there are $\binom{4}{3}$ ways to choose the places for Al, Bob, and Charlie in the top 4, $3!$ ways to arrange them in these places, and $5!$ ways to arrange the remaining 5 people in their spots;
so this gives a probability of $\displaystyle\frac{\binom{4}{3}3!5!}{8!}=\frac{4\cdot6}{6\cdot7\cdot8}=\frac{1}{14}$.
A: $4$ slots are up for grabs, so a simple way is to just  compute $\dfrac48\cdot\dfrac37\cdot\dfrac26= \dfrac{1}{14}$
