How many positive integers 'n' can be form using the digits 3, 4, 4, 5, 6, 6, 7 if we want 'n' to exceed 60,00,000?

Answer: Option C

As per the given condition, number in the highest position should be either 6 or 7, which can be done in 2 ways.
If the first digit is 6, the other digits can be arranged in $$\frac{6!}{2!}$$ = 360 ways.
If the first digit is 7, the other digits can be arranged in $$\frac{6!}{2!\times 2!}$$  = 180 ways.
Thus required possibilities for n,
= 360 + 180
= 540 ways