How many different positive, six-digit integers can be formed using the digits 2, 2, 5, 5, 9 and 9?

6 numbers for the first digit, 5 for the second digit...etc, so we have 6*5*4*3*2=6!. but, because there are two 2's, two 5's, and two 9's, we have to divide by 2!*2!*2! which is only 2^3, which is 8, so 6*5*4*3*2/8=6*5*3=$\boxed{90}$.