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?

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 number 223355888 by rearranging its digits such that the odd digits occupy even positions, we can follow these steps:

### Step 1: Identify the Digits
The number 223355888 consists of the following digits:
- Odd digits: 3, 3, 5 (total 3 odd digits)
- Even digits: 2, 2, 8, 8, 8 (total 6 even digits)

### Step 2: Determine Positions
Since we need the odd digits to occupy even positions, we first identify the even positions in a 9-digit number:
- The even positions are: 2, 4, 6, and 8 (4 positions).

### Step 3: Place the Odd Digits
We have 3 odd digits (3, 3, 5) that need to be placed in the 4 even positions. We can select any 3 out of the 4 even positions for the odd digits. The number of ways to choose 3 positions from 4 is given by the combination formula:
\[ \binom{4}{3} = 4 \]
Now, we need to arrange the 3 odd digits in the chosen positions. Since we have two 3's, the arrangement of the odd digits is given by:
\[ \frac{3!}{2!} = 3 \]
Thus, the total arrangements for placing odd digits in the even positions is:
\[ 4 \times 3 = 12 \]

### Step 4: Place the Even Digits
After placing the odd digits, we have 6 positions left (1, 3, 5, 7, 9) for the even digits (2, 2, 8, 8, 8). The arrangement of these digits is calculated as:
\[ \frac{5!}{2! \times 3!} = 10 \]
where 2! accounts for the two 2's and 3! accounts for the three 8's.

### Step 5: Calculate Total Arrangements
Finally, we multiply the arrangements of odd digits and even digits:
\[ \text{Total arrangements} = 12 \times 10 = 120 \]

Thus, the total number of different nine-digit numbers that can be formed is **120**.