In 8 digits numbers, 4 places are odd places. Also, in the given 8 digits, there are three odd digits 1,1 and 3. No. of ways three odd digits arranged at four even places =2!4P3=2!4! No. of ways the remaining five digits 2 , 2, 2, 4 and 4 arranged at remaining five places =3!2!5! Hence, required number of 8 digits number =2!4!×3!2!5!=120