# I was just wondering, if a series of numbers has 8 digits and the 1st digit is always 3, how many possible combinations of this number could there be. I am certain I once learned this in a math class years ago, but cannot remember how to calculate it. Help would be greatly appreciated. thanks

since the first digit dosn't change, you have 10 possable choices for each of the 7 variable positions, ergo, 10^7 if I understood the question.

## You're absolutely correct! To calculate the total number of possible combinations, you need to consider that the first digit is always 3, and there are 7 remaining variable positions.

Since each of these 7 positions can have 10 possible values (from 0 to 9), you can multiply 10 by itself seven times. This gives us 10^7, which equals 10,000,000.

Hence, there are 10 million possible combinations for a series of numbers with 8 digits, where the first digit is always 3.