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 4 distinct digits, there would be $4!=24$ orderings. However, we must divide by 2! once for the repetition of the digit 2, and divide by 2! for the repetition of the digit 9 (this should make sense because if the repeated digit were different we would have twice as many orderings). So, our answer is $\frac{4!}{2!\cdot 2!}=2\cdot 3=\boxed{6}$.