Question

How many numbers are there between 1245 and 5421 inclusive which contain each of the digits 1,2,4 and 5 once and once only? One of these numbers is chosen at random. Find the probability that it is
(a) divisible by 5
(b) greater than 3000

Answer 1

There are 4! = 24 numbers
12 of them start with 4 or 5 and are above 3000 -> 12/24 = 50%
6 of them end with 5 and are thus divisible by 5 -> 6/24 = 25%