Question:
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}?
Solution:
Number of ways of selecting the first term from the set {1, 2, 3} = 3
Corresponding to each of the selected first terms, the number of ways of selecting the common difference from the set {1, 2, 3, 4, 5} = 5
$\therefore$ Total number of AP's that can be formed $=3 \times 5=15$