How many numbers are there between 99 and 1000 such that 8 occupies the unit place?

Numbers between 99 and 1000 are 100 to 999, i.e. all three digit numbers. Now, for 100 to 200: 8 will come at the unit’s place 10 times, i.e. 108, 118, 128…. 198.

How many numbers are there between 99 and 1000 which have exactly one of their digits as 5?

= 900 – 648 = 252. Please mark my answer as brainliest if my answer was helpful to you . Marking my answer as brainliest will help you earn some extra points.

How many numbers are there between 100 and 1000 such that 7 is the unit place?

Hence, the total possibilities here are 90.

How many numbers are there between 99 and 1000 such that?

7 is in the unit’s place. The middle digit can be any one of the 10 digits from 0 to 9. The digit in hundred’s place can be any one of the 9 digits from 1 to 9. Therefore by the fundamental principle of counting there are 10 × 9 = 90 numbers between 99 and 1000 having 7 in the unit’s place.

How many numbers are there between 99 and 1000 having atleast one of their digits is 7?

90 numbers
The middle digit can be any one of the 10 digits from 0 to 9. The digit in hundred’s place can be any one of the 9 digits from 1 to 9. Therefore by the fundamental principle of counting there are 10 × 9 = 90 numbers between 99 and 1000 having 7 in the unit’s place.

How many numbers are there between 100 and 1000 such that every digit is either 3 or 4?

Now we will use the fact that numbers should have only 2 and 9 as their digits, so we will find all the possible arrangements that can be done using these conditions and that will be the final answer. Then the only possible number is 222.

How many numbers are there between 99 and 1000 such that the digit 8 October is the unit place?

Detailed Solution. ∴ The required result will be 90.

How many numbers are there between 1999 and 2000 such that the digit 8 occupies the unit place?

Hundred’s place can be filled up with any digit from 1–9,1–9,barring 88, i.e. 8 such digits. Units place can be filled up with any digit from 00–9,–9, barring 88 i.e. 9 such digits. This gives us 99∗8=72∗8=72numbers, with 88 in tens place. This gives us 8∗9=728∗9=72 numbers, with 88 in units place.

How many digits can occupy one’s place between 99 and 1000?

⇒ The number of digits that can occupy one’s place = 1 (only 8 can occupy this position as stated in question) ⇒ The total numbers between 99 and 1000 where unit’s place is occupied by 8 = 9 × 10 × 1 = 90 ∴ The required result will be 90.

How many times does 8 come at the same place between 99-1000?

Numbers between 99 and 1000 are 100 to 999, i.e. all three digit numbers. Now, for 100 to 200: 8 will come at the unit’s place 10 times, i.e. 108, 118, 128…. 198. Similar will be the case with number sets of 201-300, 301-400 etc. Hence, the number of times that 8 will come at unit’s place between 99 and 1000 = 10 × 9 = 90

How many numbers are there between 99 and 1000 with 9?

The last such number with 9 as unit’s place is 999. (An=999) n= (910/10)= 91. So there will be 91 numbers with 9 as unit’s place between 99 and 1000. How many numbers are there between 99 and 1000?

How many numbers are there from 0 to 9 in units place?

Units place can be filled up with any digit from 0 – 9, barring 8 i.e. 9 such digits. This gives us 9 ∗ 8 = 72 numbers, with 8 in tens place. This gives us 8 ∗ 9 = 72 numbers, with 8 in units place. This allows us have any digit from 0 – 9, except 8, in units place. This combination gives us 9 numbers