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?

1. Break Down the Problem

We need to rearrange the digits of the number 223355888 into nine-digit numbers with the condition that odd digits must occupy even positions.

Digits breakdown:

• Even digits: 2, 2, 8, 8, 8 (5 counts)
• Odd digits: 3, 3, 5, 5 (4 counts)

Positioning:

• Even positions (2, 4, 6, 8): 4 positions (we can fill these with even digits)
• Odd positions (1, 3, 5, 7, 9): 5 positions (we can fill these with odd digits)

2. Relevant Concepts

We will use the permutations formula for groups with repetitions: $\text{Permutations} = \frac{n!}{n_1! \cdot n_2! \cdot \ldots}$ Where $n$ is the total number of items, and $n_1, n_2, \ldots$ are the counts of each distinct item.

3. Analysis and Detail

Step 1: Arranging Odd Digits in Odd Positions

• We have 4 odd digits: 3, 3, 5, 5.
• The permutations of these digits can be calculated as follows: $\text{Permutations of odd digits} = \frac{4!}{2! \cdot 2!} = \frac{24}{4} = 6$

Step 2: Arranging Even Digits in Even Positions

• We have 5 even digits: 2, 2, 8, 8, 8.
• The permutations of these digits can be expressed as: $\text{Permutations of even digits} = \frac{5!}{2! \cdot 3!} = \frac{120}{2 \cdot 6} = \frac{120}{12} = 10$

4. Verify and Summarize

To find the total number of different nine-digit numbers that meet the conditions, multiply the permutations of odd and even configurations: $\text{Total arrangements} = \text{Permutations of odd digits} \times \text{Permutations of even digits} = 6 \times 10 = 60$

Final Answer

The total number of different nine-digit numbers that can be formed is 60.