how many different nine digit numbers can be formed from the number `22 33 55 888` by rearranging its digits so that the odd digits occupy even positions:

Options

A. 16
B. 36
C. 60
D. 180

Answer

To solve the problem of how many different nine-digit numbers can be formed from the digits of the number `22 33 55 888` such that the odd digits occupy the even positions, we will follow these steps:

### Step 1: Identify the digits
We have the digits: 2, 2, 3, 3, 5, 5, 8, 8, 8.

- Odd digits: 3, 3, 5, 5
- Even digits: 2, 2, 8, 8, 8

### Step 2: Determine the positions
In a nine-digit number, the positions are numbered from 1 to 9. The even positions are 2, 4, 6, and 8. We need to place the odd digits in these positions.

### Step 3: Arrange odd digits in even positions
We have 4 odd digits (3, 3, 5, 5) and 4 even positions (2, 4, 6, 8).

The number of ways to arrange these odd digits is given by the formula for permutations of multiset:
\[ \text{Number of arrangements} = \frac{n!}{n_1! \cdot n_2!} \]
where \(n\) is the total number of items to arrange, and \(n_1, n_2\) are the counts of the indistinguishable items.

Here, \(n = 4\) (total odd digits), \(n_1 = 2\) (for the two 3's), and \(n_2 = 2\) (for the two 5's):
\[ \text{Arrangements of odd digits} = \frac{4!}{2! \cdot 2!} = \frac{24}{2 \cdot 2} = 6 \]

### Step 4: Arrange even digits in odd positions
Next, we have 5 even digits (2, 2, 8, 8, 8) and 5 odd positions (1, 3, 5, 7, 9).

The number of ways to arrange these even digits is:
\[ \text{Arrangements of even digits} = \frac{5!}{2! \cdot 3!} \]
where \(n = 5\) (total even digits), \(n_1 = 2\) (for the two 2's), and \(n_2 = 3\) (for the three 8's):
\[ \text{Arrangements of even digits} = \frac{120}{2 \cdot 6} = 10 \]

### Step 5: Calculate the total arrangements
The total number of different nine-digit numbers is the product of the arrangements of odd digits and even digits:
\[ \text{Total arrangements} = \text{Arrangements of odd digits} \times \text{Arrangements of even digits} = 6 \times 10 = 60 \]

### Final Answer
Thus, the total number of different nine-digit numbers that can be formed is **60**.

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