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 5 distinct digits, there would be 5!=120 orderings. However, we must divide by 3! once for the repetition of the digit 2, and divide by 2! for the repetition of the digit 9 (note that if the repeated digits were different then we could rearrange them in that many ways). So, our answer is 5!/3! x 2!=10.
I understand now! This is the easier way, guest:)
