How many eight-bit strings contain exactly three 0's?

Up till now, I've been getting problems right and I don't know why. I've just understood the method somehow. The method of writing:

_ _ _ _ _ _ _ _

And then writing how many possible number choices on the line. For this problem, I think any 3 could have a 0. So, I let the first three have two choices, and the other five with 1 choice.

So, 2 x 2 x 2 x 1 x 1 x 1 x 1 x 1 = 8.

That doesn't look right.

Is there a proper method for probability?

Thanks!