Pages In Virtual Address Space Equation

Question

I have been given the following problem to work out..but not sure I am on the right track..

If a virtual address of 2GB is divided into 4k pages how many pages would exist in the virtual address space.

I have worked it out that if 2³¹ = 2GB and 2¹² would equal the 4k page size..

So if I take the 12 bits away from the 31 I would be left with 2¹⁹ pages..

Is that right or am I way off the mark?

Any confirmation/direction (formula) would be appreciated!