Answer
a) Possible combinations = 30^{3} = 27000
b) P(30,3) = \frac{30}{30-3} = 30*29*28 = 24360
Work Step by Step
a) Since repetition is allowed, we can choose 3 numbers in the following ways
first number can be chosen in 30 ways (any number between 1-30).
second number can be chosen in 30 ways (any number between 1-30).
third number can be chosen in 30 days(any number between 1-30).
So, total combinations = 30*30*30
b) Since repetition is not allowed, we can choose 3 numbers in the following ways
first number can be chosen in 30 ways (any number between 1-30).
second number can be chosen in 29 ways (any number between 1-30 except the first number).
third number can be chosen in 28 days(any number between 1-30 except first and second number).
So, total combinations = 30*29*28