In a relay race there are five teams A. B, C. D and E.(a)  What is the probability that A. B and C finish first, second and third, respectively.(b)  What is the probability that A. B and C are first three to finish (in any order)       (Assume that all finishing orders are equally likely).

Solution: As there are $5$ positions so total number of ways are $5!$ where $A,B,C,D\text{and}E$ finish in the race.
(a) When $A,B\text{and}C$ finish first second and third respectively, there are only $2!$ places where $D$ and $E$ can finish in the race.
So, required probability $=\frac{2!}{5!}=\frac{2}{120}=\frac{1}{60}$

(b)When $A,B,C$ are first three to finish, then $A,B,C$ can be placed in $3!$ ways.
In this case, $D,E$ can be placed in $2!$ ways.
So, total probaility in this case $=\frac{2!3!}{5!}=\frac{12}{120}=\frac{1}{10}$