In the given number ‘8467252371’ , if all the odd digits are arranged first in ascending order from left then all the even digits are arranged after that in ascending order. Find how many digits are remained same on their position after the arrangement?

The correct option is Two.

Step 1: Identify the given number and its digits
The given number is 8467252371.
Let us write down the original sequence of digits along with their positional indices (from left to right, 1 to 10):

• 1st position: 8
• 2nd position: 4
• 3rd position: 6
• 4th position: 7
• 5th position: 2
• 6th position: 5
• 7th position: 2
• 8th position: 3
• 9th position: 7
• 10th position: 1

Step 2: Separate the digits into odd and even groups
• Odd digits present in the number: 7, 5, 3, 7, 1
• Even digits present in the number: 8, 4, 6, 2, 2

Step 3: Rearrange the digits according to the given rule
1. Arrange all odd digits first in ascending order:
The odd digits in ascending order are: 1, 3, 5, 7, 7.

2. Arrange all even digits after the odd digits in ascending order:
The even digits in ascending order are: 2, 2, 4, 6, 8.

Combining both lists gives the new sequence of digits:
1, 3, 5, 7, 7, 2, 2, 4, 6, 8

Step 4: Compare positions before and after the rearrangement

• Position 1: Original = 8, New = 1 (Changed)
• Position 2: Original = 4, New = 3 (Changed)
• Position 3: Original = 6, New = 5 (Changed)
• Position 4: Original = 7, New = 7 (Same)
• Position 5: Original = 2, New = 7 (Changed)
• Position 6: Original = 5, New = 2 (Changed)
• Position 7: Original = 2, New = 2 (Same)
• Position 8: Original = 3, New = 4 (Changed)
• Position 9: Original = 7, New = 6 (Changed)
• Position 10: Original = 1, New = 8 (Changed)

Conclusion:
The digits at position 4 (which is 7) and position 7 (which is 2) remain in the exact same positions as in the original arrangement. Therefore, exactly Two digits remain unchanged in their position.