Question

John wants to form 5 digit numbers from the set of integers from 0 to 9 inclusive. Assuming he can reuse numbers, and the 5 digit number must start with and 8 and end with a 0, how many different numbers may he form?

Answer 1

`1000` possibilities

Explanation:

The first number must be an `8`, so there is only `1` possibility for the first digit.
The second number can be anywhere between `0` to `9`, so there are `10` possibilities for the second digit.
The third number can be anywhere between `0` to `9`, so there are `10` possibilities for the second digit.
The fourth number can be anywhere between `0` to `9`, so there are `10` possibilities for the second digit.
The fifth number must be a `0`, so there is only `1` possibility for the last digit.

We multiply these possibilities to get `1*10*10*10*1=1000` possibilities.