jmc

We could go ahead and count these directly, but instead we could count in general and then correct for overcounting. That is, if we had 6 distinct digits, there would be $6!=720$ orderings. However, we must divide by 2! once for the repetition of the digit 2, 2! for the repetition of the digit 5, and again 2! for the repetition of the digit 9 (this should make sense because if the repeated digits were different then we could rearrange them in 2! ways). So, our answer is $\frac{6!}{2!\cdot 2!\cdot 2!}=\boxed{90}$.