The position of how many digits in the number 3 2 1 4 6 5 9 8 7 will remain same when the digits are arranged in ascending order?

Options

A. None
B. one
C. Two
D. Three

Explanation

To solve the problem of how many digits in the number 321465987 will remain in the same position when the digits are arranged in ascending order, we can follow these steps:

### Step 1: Write down the original number.
The original number is: **321465987**.

### Step 2: Arrange the digits in ascending order.
To arrange the digits in ascending order, we need to sort them from the smallest to the largest. The sorted order of the digits is: **123456789**.

### Step 3: Compare the original and sorted sequences.
Now, we will compare the original number with the sorted number to see which digits remain in the same position.

- **Original**: 3 2 1 4 6 5 9 8 7
- **Sorted**: 1 2 3 4 5 6 7 8 9

### Step 4: Identify the positions that remain the same.
We will check each position (1 to 9) to see if the digit in the original number matches the digit in the sorted number:

1. Position 1: 3 (original) vs 1 (sorted) - Not the same
2. Position 2: 2 (original) vs 2 (sorted) - Same
3. Position 3: 1 (original) vs 3 (sorted) - Not the same
4. Position 4: 4 (original) vs 4 (sorted) - Same
5. Position 5: 6 (original) vs 5 (sorted) - Not the same
6. Position 6: 5 (original) vs 6 (sorted) - Not the same
7. Position 7: 9 (original) vs 7 (sorted) - Not the same
8. Position 8: 8 (original) vs 8 (sorted) - Same
9. Position 9: 7 (original) vs 9 (sorted) - Not the same

### Step 5: Count the digits that remain in the same position.
From the comparison, the digits that remain in the same position are:
- Position 2: 2
- Position 4: 4
- Position 8: 8

So, there are **3 digits** that remain in the same position.

### Final Answer:
The answer is **3**.

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