How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions?163660180

Sure, here is the step by step solution:

Step 1: Identify the odd and even digits in the number 223355888. The odd digits are 3, 3, 5, 5 and the even digits are 2, 2, 8, 8, 8.

Step 2: There are 9 positions in the number and we want the odd digits to occupy the even positions. In a 9 digit number, there are 4 even positions (2nd, 4th, 6th, 8th) and 5 odd positions (1st, 3rd, 5th, 7th, 9th).

Step 3: We have 4 odd digits (3, 3, 5, 5) and 4 even positions. The number of ways to arrange these 4 odd digits in 4 even positions is 4!/2!2! = 6. (We divide by 2! twice because the digit 3 and 5 are repeated twice).

Step 4: We have 5 even digits (2, 2, 8, 8, 8) and 5 odd positions. The number of ways to arrange these 5 even digits in 5 odd positions is 5!/2!3! = 10. (We divide by 2! once for the two 2s and 3! once for the three 8s).

Step 5: Therefore, the total number of ways to arrange the digits is 6 * 10 = 60.

So, there are 60 different nine digit numbers that can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions.