How many different nine digit numbers can be formed from the number 22 33 55 8 88 by rearranging its digits so that the odd digits occupy even positions?

The four digits 3, 3, 5, 5, can be arranged at four even places in 4!2!2!=6 ways and the remaining digits viz., 2, 2, 8, 8, 8 can be arranged at the five odd places in 5!2!3!=10 ways.Thus, the number of possible arrangements is (6) (10) = 60.