**Concept: The question tests on the area of Modern Math(Counting/Combinatorics)**

**Solution:**

2310 broken into its factors = 1 * 2 * 3 * 5 * 7 * 11

Let us take the 5 factors 2,3,5,7 and 11 and find the number of sets possible

Taking 2 - 2 - 1 = 5C2 * 3C2 * 1C1 / 2! = 15 ways

Taking 1 - 1 - 3 = 5C1 * 4C1 * 3C3 / 2! = 10 ways

Taking 2 - 3 and the last number being 1. 5C2 * 3C3 = 10 ways

Taking 4 - 1 and the last number being 1. 5C4 * 1C1 = 5 ways

Total number of ways = 15 + 10 + 10 + 5 = 40 ways

**(Option D)**