How many A.P.’s with 10 terms are there whose first term is in the set {1, 2, 3} and whose common difference is in the set {1, 2, 3, 4, 5}?
Each AP consists of a unique first term and common difference.
Number of ways to select the first term of a given set is 3C1 = 3
Number of ways to select a common difference of given set is 5C1 = 5
Hence total number of AP’s possible are 3C1 × 5C1 = 3 × 5 = 15