How many different nine digit numbers can be formed from the numebr 223377888 by rearranging its digits so that odd digits occupy even places

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 223377888 with the condition that odd digits occupy even places, we can follow these steps:

### Step 1: Identify the digits and their counts
We have the digits: 2, 2, 3, 3, 7, 7, 8, 8, 8.
- Even digits: 2, 2, 8, 8, 8 (5 even digits)
- Odd digits: 3, 3, 7, 7 (4 odd digits)

### Step 2: Determine the positions
In a nine-digit number, the positions are labeled as follows:
1. 1st (odd)
2. 2nd (even)
3. 3rd (odd)
4. 4th (even)
5. 5th (odd)
6. 6th (even)
7. 7th (odd)
8. 8th (even)
9. 9th (odd)

According to the problem, we need to place the odd digits in the even positions, which are 2nd, 4th, 6th, and 8th.

### Step 3: Place the odd digits in the even positions
We have 4 odd digits (3, 3, 7, 7) to place in 4 even positions. Since there are repetitions (two 3s and two 7s), we calculate the arrangements as follows:

\[ \text{Arrangements of odd digits} = \frac{4!}{2! \times 2!} = \frac{24}{4} = 6 \]

### Step 4: Place the even digits in the remaining positions
Now we have to fill the remaining 5 positions (1st, 3rd, 5th, 7th, and 9th) with the even digits (2, 2, 8, 8, 8). Again, we have repetitions (two 2s and three 8s), so we calculate the arrangements as follows:

\[ \text{Arrangements of even digits} = \frac{5!}{2! \times 3!} = \frac{120}{12} = 10 \]

### Step 5: Calculate the total arrangements
The total number of different nine-digit numbers can be calculated by multiplying the arrangements of odd digits and even digits:

\[ \text{Total arrangements} = \text{Arrangements of odd digits} \times \text{Arrangements of even digits} = 6 \times 10 = 60 \]

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

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