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).

Explanation

As there are `5` positions so total number of ways are `5!` where `A,B,C,D and E` finish in the race.

(a) When `A,B 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 `= (2!)/(5!) = 2/120 = 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 `= (2!3!)/(5!) = 12/120 = 1/10`