Hello! This is a classic example of a factorial problem. We have been told that no number can be used more than once, but order isn't mentioned, so we will assume order doesn't matter, which you can easily write as:
nPr = n! / (n-r)!
In this case, "n" is the number of things to choose from and "r" is the number of things you are picking from "n" objects.
Here, we have n=6 (things to choose from) and r=4 (things we are picking from them. We have:
6! / (6-4)! = (1 x 2 x 3 x 4 x 5 x 6) / (1 x 2) = 3 x 4 x 5 x 6 = 12 x 30 = 360 possible combinations.
Part 2 of your question:
Here, we know that our first digit must be 8 or 9. Luckily, this means that we can have any combination of {1,2,3,5 and (8 or 9)} depending on the first digit. So, we can have 9, followed by any random, non-repetitive 3-digit number of the set {1,2,3,5,8}, and an identical number of combinations for 8 followed by a 3-digit combination of {1,2,3,5,9}. This means we can simply calculate the number of combinations for one instance and double it.
For example, assume your first digit is 9 :
n = 5, r = 3... we can right this as 5P3 = 5! / (5-3)! = 3 x 4 x 5 = 60 combinations.
This would be identical if the first digit is 8, so we can double this result to get our final answer.
The answer to part 2 is 120 possible numbers.
