How many positive integers ‘n’ can we form using the digits 3, 4, 4, 5, 6, 6, 7 if we want ‘n’ to exceed 6,000,000?

We are given exactly 7 digits - 3, 4, 4, 5, 6, 6, 7. The millions digit can be either 6 or 7. 

Case 1: When the millions digit is 6. 

6 _ _ _ _ _ _ We are left with six digits {3, 4, 4, 5, 6, 7}. 

These six digits can be arranged in six places in $$\dfrac{6!}{2!}$$ ways.

Case 1: When the millions digit is 7. 

7 _ _ _ _ _ _ We are left with six digits {3, 4, 4, 5, 6, 6}. 

These six digits can be arranged in six places in $$\dfrac{6!}{2!*2!}$$ ways.

Therefore, total number of numbers 'n' = $$\dfrac{6!}{2!}$$ + $$\dfrac{6!}{2!*2!}$$ = 360 + 180 = 540. Hence, option C is the correct answer.