Let A be the set of all prime factors of 2310 and f: A → B be the function given by f(x) = log₂(x² + [x³/3]).

where B is the range of f(x), [t] is G.I.F. The number of one-one functions are:

• A: 20
• B: 36
• C: 120
• D: 720

To solve the problem, we first need to determine the prime factors of 2310. The prime factorization of 2310 is:

• 2310 = 2 × 3 × 5 × 7 × 11

This means the set A of all prime factors is:

• A = {2, 3, 5, 7, 11}

The size of set A is 5, as it contains 5 elements. Next, we need to find the range of the function f(x) = log₂(x² + [x³/3]). The greatest integer function [t] rounds down to the nearest integer.

Since the function f maps each element of A to a unique value in B, we can determine the number of one-to-one functions from A to B. The number of one-to-one functions from a set of size m to a set of size n is given by:

n! / (n - m)!

In this case, we have:

• m = 5 (the size of set A)
• n = 5 (assuming the range B also has at least 5 distinct values)

Thus, the number of one-to-one functions is:

5! / (5 - 5)! = 5! / 0! = 120 / 1 = 120

Therefore, the answer is:

C: 120