How many numbers are there in between 100 and 10000, such that 5 is in the unit place?

Hint: We need to first understand the given conditions of numbers being inside 100 and 1000 and 5 being its unit digit. Then we apply the concept of choices to fill-up places for the hundredth place and the tenth place. We multiply the number of choices to find the final answer.

Complete step by step answer:
All the numbers in the range of 100 to 1000 are of three digits. So, we need to find three-digit numbers which have 5 as its unit digit.
Here we apply the concept of permutation and combination.
We need to fill-up three places out of which 1 place is fixed with a fixed number. The remaining number of places is 2. The choices to fill up the places are 0 to 9.
The hundredth place can’t be 0 as then the whole digit becomes a two-digit number from a three-digit number.
So, basically, for the hundredth place, we have choices of 1 to 9 and for the tenth place, we have choices of 0 to 9.
For the hundredth place, we have 9 choices and for the tenth place, we have 10 choices.
The unit place is fixed with 5.
So, the number of ways these numbers can be created is $9\times 10=90$.
Therefore, there are 90 such numbers.

Note:
 We know that the unit place is fixed. So, there we don't have any choices to change. We could have solved it by taking the complement set of unit place being any number but 5 and subtracted that from a total number of three-digit numbers.