According to the question,
1,1,1,2,2,3,3,4,4
there are four even digits as 2 and 4 are repeating two times each.
So, the even digits will occupy the even places in (2!2!)4!=6 ways
3 and 1 are repeating two and three times respectively.
So, the remaining places will be occupied by odd digits in (3!2!)5!=10 ways
Therefore, total 9 digits that can be formed =6×10=60.
Hence, the answer is 60.